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From: sanboz on 17 Nov 2009 16:01

"Mark Murray" <w.h.oami(a)example.com> wrote in message
news:4b0251ab$0$2482$db0fefd9(a)news.zen.co.uk...
> sanboz wrote:
>> Any Ideas?
>
> Use something else? Thunderbird?
>
> M

I'll check it out with some others, see what happens


From: Peter on 17 Nov 2009 18:37
On Nov 15, 10:32 pm, JSH <jst...(a)gmail.com> wrote:
>
> I have read up a lot on issues in this area.  

Stop the lying. You don't read math books for 3 simple reasons:
1) You ignorant
2) You're arrogant
3) You're immature and don't have the perseverance.


HTH,

M
From: Mark Murray on 28 Nov 2009 17:37
JSH wrote:
Exhibit a)
> If that is in your code I missed it. Oh, but it is amazing to me how
> compact what you did present is!

Exhibit b)
> And you say it actually got the right answer at times? If so, and you
> DID leave out a key piece that could be of research interest as to
> why.

Exhibit c)
> Good work though with what you have. Can you give an example where it
> WORKS as well. I find it intriguing that it works at all.

Exhibit d)
> If it DOES sometimes work, it still might be of some interest even if
> you DID use the full algorithm, and I just didn't see where.

For something that you PROVED to work, your lack of knowledge of YOUR
TSP algorithm is astonishingly deficient.

What is amusing is the apparent surprise you show that it actually
works at all. Some proof you have there.

And this is what you were going to rescue the British Empire with?

HAH! Some guy called Dijkstra will have to do for now.

M
From: spudnik on 28 Nov 2009 18:11
well, one of HSJ's gedanken experiments bore fruit
with your own programming; makes me want almost
to get into it, myself. I always thought it amuzing,
though others aver that it makes no difference, that
the curvature of Earth (i.e. space) isn't considered as a part
of the problem -- truly, it is generally devized as a pure
application of planar graphtheory, which includes spherical
as a special case, I hear (more elementary stuff to grok),
just as the four-color theorem was turned into that,
in order to save on ink-coloration-by-hand.

now, on the other hand, going beyond the modality
of zero-sum-games with the nodal overview
of both of the could-be salesforce
-- thanks to John "Memorial Nobelist-economist" Nash --
such a communicative approach would be required, if
it was Two Travelling Salesfolk Orienteering (which is
just another version of my new game, EGO,
for one or more players .-)

> As an experiment, I also tried setting the distances equal to the
> costs instead, but to my surprise that worked much less often than the
> distance normalized version. Though in retrospect it's fairly obvious
> why; much of the time, having the two travellers seek to stay close to
> one another is actually a bad idea. Consider nodes arranged in a
> circle with pairs of points close to one another, but in which the
> pairs themselves are far apart, for example.

--go l'OEuvre!
http://www.21stcenturysciencetech.com/Articles_2009/Relativistic_Moon.pdf
http://wlym.com/~animations/fermat/index.html

thus quoth:
(1) Ampère's demonstration of the physical presence of an angular
force, essentially overthrowing the fundamental assumption of
potential theory as still taught, and its conclusive experimental
proof by the 10-year collaboration of Carl Friedrich Gauss and Wilhelm
Weber;
(2) The 1855 Weber-Kohlrausch experiment, establishing the relative
velocity at which the force between electrical particles is reduced to
zero, and provoking Bernhard Riemann to propose (1858) a similarity in
the propagation of light and the electrodynamic potential;
(3) Weber's subsequent deduction (1871) of the bound state of pairs of
like-charged particle/waves within the confines of a 10-16 to 10-13 cm
spherical radius, establishing the natural basis for the formation of
the atomic nucleus.
In the period from 1999 to 2006, I was able to apply that
understanding of the Ampere-Gauss-Weber electrodynamics to the
Keplerian model of the atomic nucleus proposed in 1985 by Dr. Moon.2 I
arrived at a structure which at once overcame what had been two of the
leading objections to the Rutherford-Bohr-Sommerfeld model of the
atom, without the need to invoke any new conditions ad hoc. The
objections of leading chemists, Lewis, Parsons, Langmuir and others,
to the Bohr atom were summarized by
From: JSH on 29 Nov 2009 10:22
On Nov 28, 5:00 pm, Rotwang <sg...(a)hotmail.co.uk> wrote:
> Sorry if this appears twice, my internet connection keeps cutting out.
>
> On 28 Nov, 22:58, JSH <jst...(a)gmail.com> wrote:
>
>
>
>
>
> > On Nov 28, 10:55 am, Rotwang <sg...(a)hotmail.co.uk> wrote:
>
> > [...]
>
> > > It isn't, but that wasn't a mistake on my part: as I said, I coded
> > > what you called the "distance normalized" version of the algorithm, in
> > > which the distance between any two distinct nodes is simply set to
> > > one. Quoting from your blog:
>
> > Ok.  That makes sense now.
>
> > >   After pondering the problem I have a distance normalized TSP
> > >   algorithm, which simply assumes that the distance between every
> > >   node is a unit of 1, so with this algorithm, with m nodes, the
> > >   nodes are assumed to be in an m-1 dimensional space with a
> > >   distance of 1 from each other in that space.
>
> > > As an experiment, I also tried setting the distances equal to the
> > > costs instead, but to my surprise that worked much less often than the
> > > distance normalized version. Though in retrospect it's fairly obvious
> > > why; much of the time, having the two travellers seek to stay close to
> > > one another is actually a bad idea. Consider nodes arranged in a
> > > circle with pairs of points close to one another, but in which the
> > > pairs themselves are far apart, for example.
>
> > > > Oh, but it is amazing to me how
> > > > compact what you did present is!
>
> > > > And you say it actually got the right answer at times?
>
> > > Of course, but then an algorithm that simply returned the nodes in a
> > > random order would also get the right answer at times. In fact, your
>
> > Well yeah, that's not a surprise.
>
> > > algorithm gave the correct answer for most inputs, though the
> > > proportion it got right decreased with the number of nodes (e.g. only
> > > about 30 failures per 1000 attempts with 5 nodes, about 170 failures
>
> > Yuck.  That doesn't sound bad.  With 5 nodes, 5! = 120 possible brute
> > force paths, with 5*5 chances to match the algorithm as it is O(m^2),
> > and you have 25/120 or 20.8% success rate for random.
>
> > Which would give about 800 failures per 1000 attempts.
>
> This calculation doesn't make much sense, as far as I can tell.
> Firstly, assuming that a graph has no unusual symmetries, the
> probability of a random guess giving the optimal path is (where m is
> the number of nodes) 2/(m - 1)!. This is because the length of a path

Oh, yeah, you're right. Read over the rest and I find it convincing
to the position that my idea is no better than the greedy algorithm.

So I didn't solve the TSP problem and did not prove P=NP. Oh well.
Not a big deal.


James Harris
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