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Nov 30, 2016 at 6:38 comment added j0equ1nn Maybe I'm beating a dead horse now, but I'd say this question is frivolous if and only if the result that the graph is non-planar is frivolous.
Nov 27, 2016 at 21:54 history edited j0equ1nn
added soft-question tag
Nov 27, 2016 at 21:51 comment added j0equ1nn @DavidWhite The math involved in answering this may not be "research level," but I thought some users of this site would share my curiosity in answering it since it pertains to our community (and some did). I don't think that personally finding a question frivolous is sufficient for rejecting it as off-topic, but I do think I should have marked it as a "soft question," which I've now done.
Nov 27, 2016 at 21:34 history closed David White
Alex Degtyarev
Wolfgang
Stefan Kohl
Chris Godsil
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Nov 27, 2016 at 15:55 review Close votes
Nov 27, 2016 at 21:34
Nov 27, 2016 at 15:34 comment added David White This seems like a frivolous question to me.
Nov 27, 2016 at 1:52 answer added Pat Devlin timeline score: 2
Nov 27, 2016 at 1:17 comment added j0equ1nn @PatDevlin Yeah, that's how I felt at first too. Like, saying that the graph is non-planar is basically just saying there has been an instance of multiple advisors in a way that caused it, but this isn't how the analysis is presented. I'd like to see it put that way.
Nov 27, 2016 at 1:14 history edited j0equ1nn CC BY-SA 3.0
boldfaced main question
Nov 27, 2016 at 1:13 comment added Pat Devlin The thing is probably very close to a forest, as very few people have more than one advisor.
Nov 27, 2016 at 1:10 comment added j0equ1nn @PatDevlin I wonder if that data is available, though the data I suggest attaining is more specific to the nature of this graph.
Nov 27, 2016 at 1:09 history edited j0equ1nn CC BY-SA 3.0
boldfaced main question
Nov 27, 2016 at 0:47 comment added Pat Devlin A measure of non-planarity is crossing number, which is rather well controlled in terms of the number of edges.
Nov 27, 2016 at 0:35 history asked j0equ1nn CC BY-SA 3.0