Timeline for Is the value of $\max_{f:V\rightarrow [\frac{-1}{2},\frac{1}{2}], \\ \sum_{v}{f(v)}=0} \frac{f^T L_G f}{n-f^Tf}$ polynomially computable?

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Dec 21, 2016 at 22:54	history	bumped	CommunityBot		This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Nov 22, 2016 at 16:37	comment	added	Igor Shinkar		Does it have any meaning ? geometric or combinatorial?
Nov 21, 2016 at 22:03	comment	added	j.s.		@Sasho Nikolov. you are right. I apologize for my carelessness. I edit my question.
Nov 21, 2016 at 22:01	history	edited	j.s.	CC BY-SA 3.0	added 34 characters in body; edited title
Nov 21, 2016 at 20:45	comment	added	Sasho Nikolov		you are right. but your formulation still allows a zero denominator. say, $n$ is even, and $f(v) \in \{\pm 1\}$ is positive on half the vertices.
Nov 21, 2016 at 20:38	history	edited	j.s.	CC BY-SA 3.0	deleted 27 characters in body
Nov 21, 2016 at 19:40	comment	added	j.s.		@Sasho Nikolov. for the fraction in the remark, contant part of $f$ gives value $0$ in the numerator and a nonnegetive value in denominator.
Nov 21, 2016 at 19:32	history	edited	j.s.	CC BY-SA 3.0	added 11 characters in body
Nov 21, 2016 at 18:33	answer	added	Klaus Draeger		timeline score: 1
Nov 21, 2016 at 16:58	history	edited	j.s.	CC BY-SA 3.0	added 247 characters in body
Nov 20, 2016 at 21:33	comment	added	Sasho Nikolov		The denominator could be zero in any dimension, as there is no integrality constraint, it seems.
Nov 20, 2016 at 18:14	comment	added	Klaus Draeger		Also note that in dimension $4$ and above, there will be vectors for which your denominator is $0$ due to the four-square theorem.
Nov 20, 2016 at 18:12	comment	added	Klaus Draeger		$x$ is an integer vector? Any other constraints?
Nov 20, 2016 at 17:12	history	asked	j.s.	CC BY-SA 3.0