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Sep 27, 2012 at 15:19 vote accept Craig Feinstein
Sep 21, 2010 at 12:55 comment added Craig Feinstein Gerry, see rjlipton.wordpress.com/2009/02/17/…
Sep 21, 2010 at 3:56 comment added Gerry Myerson Craig, can you tell us more than "a theorem by Dobkin and Lipton"? Say, the title of a paper?
Sep 21, 2010 at 2:27 answer added Tracy Hall timeline score: 9
Sep 20, 2010 at 15:25 comment added Tracy Hall @David Speyer: Don't forget hyperplanes like $a_1=a_3$. (For uniqueness, either the sets or their complements can be taken to be disjoint, but not necessarily both.)
Sep 20, 2010 at 14:24 comment added Tracy Hall One simplifying observation: If you make the additional assumption that the $a_i$ are all positive, it reduces the number of sortings by a factor of exactly $2^n$. (Replacing any one number by its negative gives the same ordering under complementation with respect to that one index.)
Sep 20, 2010 at 14:24 comment added Craig Feinstein I am guessing that the answer is $2^{\Theta(n^2/2)}$, because of a theorem by Dobkin and Lipton.
Sep 20, 2010 at 13:53 comment added David E Speyer Nice question. One way to phrase it is to ask how many regions $\mathbb{R}^n$ is divided into by the hyperplanes of the form $\sum_{i \in I} a_i = \sum_{j \not \in I} a_j$. I feel like this example should be in some survey on combinatorics of hyperplane arrangements, but I don't know which one.
Sep 20, 2010 at 13:47 history edited David E Speyer CC BY-SA 2.5
Fixed Latex
Sep 20, 2010 at 13:45 comment added Craig Feinstein I tried to put all 8 sortings of S_2, but it stopped me. Here they are: {0<a_1<a_2<a_1+a_2, 0<a_2<a_1<a_1+a_2, a_2<0<a_1+a_2<a_1, a_1<0<a_1+a_2<a_2, a_2<a_1+a_2<0<a_1, a_1<a_1+a_2<0<a_2, a_1+a_2<a_1<a_2<0, a_1+a_2<a_2<a_1<0.
Sep 20, 2010 at 13:39 history asked Craig Feinstein CC BY-SA 2.5