Skip to main content
17 events
when toggle format what by license comment
Apr 27, 2015 at 1:04 comment added Craig Feinstein It seems to me that the problem here is precisely the negative weights. My proposed Monte Carlo simulation doesn't solve the problem.
Apr 27, 2015 at 0:50 vote accept Craig Feinstein
Apr 27, 2015 at 0:49 vote accept Craig Feinstein
Apr 27, 2015 at 0:49
Apr 26, 2015 at 18:23 comment added Hari Krovi You might want to look at these papers. arxiv.org/abs/quant-ph/0611156 and arxiv.org/abs/quant-ph/0611241
Apr 26, 2015 at 16:29 comment added Craig Feinstein @CarloBeenakker thank you, I'll check that paper out.
Apr 26, 2015 at 15:39 answer added Will Sawin timeline score: 17
Apr 26, 2015 at 15:38 comment added Carlo Beenakker the problem you are running into with the negative weight in your Monte Carlo algorithm is the infamous "negative sign problem"; it is believed to be NP-hard --- arxiv.org/abs/cond-mat/0408370
Apr 26, 2015 at 15:34 history edited Craig Feinstein CC BY-SA 3.0
added word "fundamentally"
Apr 26, 2015 at 15:25 history edited Craig Feinstein CC BY-SA 3.0
added comment about zeb's comment
Apr 26, 2015 at 15:18 comment added Craig Feinstein @Zeb, in general yes. But when looking at Shor's original paper arxiv.org/abs/quant-ph/9508027, the quantum gates that he uses for the Fourier transform are simple enough that perhaps normalizing the columns in the new matrices will not mess things up too badly.
Apr 26, 2015 at 10:09 comment added zeb Wait - "normalize the columns of the new matrix"? Doesn't this step completely mess up the algorithm?
Apr 26, 2015 at 9:24 comment added user9072 It can be easier to typeset matrices using the command that exists to that end $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$ $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$
Apr 26, 2015 at 9:23 history edited user9072 CC BY-SA 3.0
fixed broken MJ
Apr 26, 2015 at 7:25 history edited Ricardo Andrade CC BY-SA 3.0
replaced new tag with existing ones; added relevant tags
Apr 26, 2015 at 7:18 history edited Ricardo Andrade CC BY-SA 3.0
replaced new tag with existing ones; added relevant tags
Apr 26, 2015 at 6:42 comment added Geoffrey Irving Any boolean circuit can be turned into a Bayesian network, so if MCMC always works then P = NP.
Apr 26, 2015 at 5:12 history asked Craig Feinstein CC BY-SA 3.0