Timeline for Symmetry preserving graph products
Current License: CC BY-SA 3.0
12 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| S Jan 16, 2018 at 18:37 | history | suggested | jeq | CC BY-SA 3.0 |
Copied image to imgur.com, as it was not being displayed because of the new https rule. Added link to original image source.
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| Jan 16, 2018 at 17:23 | review | Suggested edits | |||
| S Jan 16, 2018 at 18:37 | |||||
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Jul 20, 2012 at 22:12 | vote | accept | Hans-Peter Stricker | ||
| Jul 20, 2012 at 21:45 | answer | added | W Edwin Clark | timeline score: 2 | |
| Jul 20, 2012 at 16:23 | history | edited | Hans-Peter Stricker | CC BY-SA 3.0 |
added 338 characters in body
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| Jul 20, 2012 at 7:44 | comment | added | Hans-Peter Stricker | Aaron's example $K_6$ minus a matching has degree 4 which is more than half the number of vertices. Did I misunderstand your argument? | |
| Jul 20, 2012 at 7:31 | comment | added | Zsbán Ambrus | Also, the complement of any transitive graph is transitive. The case when one of the factors is $ K_2 $ might thus require special examination, because that's the only way a product can have a degree near half the number of vertexes. | |
| Jul 20, 2012 at 7:30 | comment | added | Zsbán Ambrus | Such a product (unless it's trivial) must have a degree at most half of the number of vertexes. There are transitive graphs of almost any number of vertexes and degree (odd number of vertexes and odd degree is forbidden): in fact, there's a circulant with any degree (a circulant is a graph with an automorphism that is a cyclic permutation moving all vertices). | |
| Jul 20, 2012 at 6:38 | answer | added | Aaron Meyerowitz | timeline score: 2 | |
| Jul 20, 2012 at 6:03 | history | asked | Hans-Peter Stricker | CC BY-SA 3.0 |