home *** CD-ROM | disk | FTP | other *** search
- Path: sparky!uunet!munnari.oz.au!manuel.anu.edu.au!dubhe.anu.edu.au!nunki.anu.edu.au!not-for-mail
- From: jmr@cs.anu.edu.au (Mike Robson)
- Newsgroups: comp.theory
- Subject: Re: P=NP; the proof
- Date: 8 Jan 1993 16:22:40 +1100
- Organization: Computer Science Department, ANU, Australia
- Lines: 42
- Distribution: inet
- Message-ID: <1ij330INN2h8@nunki.anu.edu.au>
- References: <1idr1nINNrl@nunki.anu.edu.au> <1ifcmpINN11k@nunki.anu.edu.au> <1ifljqINN137@nunki.anu.edu.au> <24567@alice.att.com>
- NNTP-Posting-Host: nunki.anu.edu.au
- Keywords: Collapse of hierarchy
-
- reeds@alice.att.com (Jim Reeds) writes:
-
-
-
- >jmr@cs.anu.edu.au (Mike Robson) writes:
-
- >>[...]
- >>I'll go back to trying to decide if the current Gismondi and Swart
- >>linear programming formulation has the required property (that every
- >>solution covers a permutation). If that claim is true, then I think
- >>my approach is a good way to try and prove it.
-
- >This is rather more than G&S themselves state. If every point in the
- >feasible region covers a permutation, then the extreme points of the
- >feasible region are exactly the permutations, and hence every point in
- >the feasible region is expressible as a convex combination of permutations.
-
- I should have been more precise. I think they are claiming that every
- solution has covers the s and t variables of a permutation (not the u).
- I think that gets round your argument; if not things look bad.
-
- >But all G&S claim is that any solution is expresible as an affine
- >combination of permutations, with AT MOST ONE NEGATIVE COEFFICIENT.
-
- Where did they claim that?
-
- >(I would like to see a real proof that any solution is expressible
- >as an affine combination of permutations, regardless of sign. Equivalently,
- >I'd like to know the dimension of the affine hull of the G&S feasible
- >region.)
-
- They claim to have a complete proof based on the dimensions of the spaces
- involved but I haven't seen any details.
- I have an idea of how to construct a proof by taking an arbitrary
- solution and systematically zeroing out all u variables except those
- of the identity permutation vector. This can be done by subtracting
- carefully chosen permutations, possibly preceded by adding other
- permutations which don't involve any u variables except the ones
- which you haven't got round to yet.
- >Jim
-
- Mike.
-