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May 3, 2018 at 18:48 comment added Suvrit @RodrigodeAzevedo -- the OP changed the question's statement to make it not a QP; the original version as stated was a convex QP.
May 3, 2018 at 15:37 vote accept Turbo
S May 3, 2018 at 15:08 history suggested Rodrigo de Azevedo CC BY-SA 4.0
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May 3, 2018 at 13:42 comment added Turbo @RodrigodeAzevedo yes QCLP.
May 3, 2018 at 13:32 review Suggested edits
S May 3, 2018 at 15:08
May 3, 2018 at 6:40 answer added Michal Adamaszek timeline score: 3
May 3, 2018 at 3:43 history edited Turbo CC BY-SA 4.0
edited title
May 3, 2018 at 2:26 comment added Turbo @Suvrit Only $x,y\in\mathbb R$.
May 3, 2018 at 2:25 history edited Turbo CC BY-SA 4.0
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May 3, 2018 at 2:23 comment added Suvrit The moment you have equality, things break down, because then you essentially have a non-convex constraint quadratic constraint, and this problem seems to not turn into an eigenvalue problem; this is not yet a proof of NP-Hardness (that's of course used a little loosely here, given that this is over real numbers not integers, but that's ok).
May 3, 2018 at 1:54 comment added Turbo @Suvrit Thank you. $z$ contains $x_i$, $y$ and other intermediate variables and in my case $xQx'=y$ holds (not $xQx'\leq y$) and you assign $y$ exactly $xQx'$ and I am not completely sure.
May 3, 2018 at 1:53 history edited Turbo CC BY-SA 4.0
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May 3, 2018 at 1:51 comment added Suvrit It depends on what 'z' contains, because otherwise it's just a convex optimization problem.
May 3, 2018 at 1:41 history edited Turbo CC BY-SA 4.0
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May 3, 2018 at 1:22 history edited Turbo CC BY-SA 4.0
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May 3, 2018 at 1:16 history asked Turbo CC BY-SA 4.0