JSH: Some history is the future
JSH
research is the possibility of continual transmission of information
from the future to the past in order to CREATE the future, where key
is what I call the optimal path algorithm.
Used against the Traveling Salesman Problem it gives you a traveler
going backwards in time to meet himself, where the algorithm requires
continual communication between the two travelers in order to get the
optimal path.
If that is a routine part of nature then light takes the optimal path
in that way, and it also gives an arrow to time--we think we're
traveling forward in the future as we're the collapsed path, when
actually we're traveling both forwards AND backwards in time.
The collapse to an optimal path gives us the illusion of only going
forwards in time.
The arrow is the collapsed optimal path which appears to only go
forward in time.
If so, then some of our "history" can be information transmitted to
the past in order to create our future (and our present).
And that includes legends and mythologies, so yes, Revelations, for
instance, seems to have future weapons in it. With this theory that's
because it actually hasn't happened yet, but was information sent to
the past about what will happen in order to MAKE it happen.
Parts of Genesis appear also to actually be stories about the future,
not the past.
So the future is an active participant in creating itself.
Which means there is no way to know what knowledge is future knowledge
transmitted to the past, unless that future wishes you to know, and
then of course you CAN know. And there seem to be no limits. (So
yes, conceivably you can get winning lottery numbers from the future,
but ONLY if the future needs you to have that information to create
itself.)
And if you can understand all of that you gotta be a singular type of
human being as it is very confusing. But one thing is clear, if I'm
talking about it, and I'm right, then the only reason I know to even
tell it is because the future needs me to know and needs me to tell it
for that future to exist.
It seems to me that such mind-numbing information would have a
predictably large impact on a human population, which is why people
haven't maybe really known it before; therefore, it seems likely some
epic event is about to occur, which is allowing the information to be
known.
For various reasons the idea is floating around that the "end of the
world" is in 2012. My memory is that Sir Isaac Newton actually
calculated the correct year and he got 2010, but I've seen no mention
of that in the record, so I'm not sure why I have that number.
It may be future knowledge.
James Harris
7
Totally vague argument.
If only you could send a lottery win number into the past in a clear
and useful way to prove it, then you would. Sending a 16 digit number
is nothing compared to sending back chapters for a book.
sanboz
"JSH" <jst...@gmail.com> wrote in message
news:04874ca3-d17f-4100...@h40g2000prf.googlegroups.com...
> One of the weirder things that has emerged from my mathematical
> research is the possibility of continual transmission of information
> from the future to the past in order to CREATE the future, where key
> is what I call the optimal path algorithm.
Join the Ashtar Command, they do that all the time. They have "channelers"
that get furure information from Aliens in Spaceeships waiting to land, and
beam it to Earth where it is used for profit by the Channelers.
I am sure the path is sub-optimal, as they (Aliens) are not here yet,
although they can talk to us.
>
> Used against the Traveling Salesman Problem it gives you a traveler
> going backwards in time to meet himself, where the algorithm requires
> continual communication between the two travelers in order to get the
> optimal path.
What if he does it twice, then there would be 4 ?
that would be a 2 ^ k problem, and 2 ^ k different paths
>
> If that is a routine part of nature then light takes the optimal path
> in that way, and it also gives an arrow to time--we think we're
> traveling forward in the future as we're the collapsed path, when
> actually we're traveling both forwards AND backwards in time.
arrow OF time. not "to"
>
> The collapse to an optimal path gives us the illusion of only going
> forwards in time.
and some of us stay stuck, with illusion of being sucessful
>
> The arrow is the collapsed optimal path which appears to only go
> forward in time.
>
> If so, then some of our "history" can be information transmitted to
> the past in order to create our future (and our present).
but history is created by the present, and is already over in the past.
>
> And that includes legends and mythologies, so yes, Revelations, for
> instance, seems to have future weapons in it.
It does not, read it, you have been watching too much TV
> With this theory that's
> because it actually hasn't happened yet, but was information sent to
> the past about what will happen in order to MAKE it happen.
>
> Parts of Genesis appear also to actually be stories about the future,
> not the past.
appear, stories ? sounds like fiction, dude
>
> So the future is an active participant in creating itself.
circular thinking, what gave you that idea, "itself" ??
>
> Which means there is no way to know what knowledge is future knowledge
> transmitted to the past, unless that future wishes you to know, and
> then of course you CAN know. And there seem to be no limits. (So
> yes, conceivably you can get winning lottery numbers from the future,
> but ONLY if the future needs you to have that information to create
> itself.)
Therefore, you must accept the fact that your research is still meaningless
into the future, because it is not sucessful in the past
>
> And if you can understand all of that you gotta be a singular type of
> human being as it is very confusing. But one thing is clear, if I'm
> talking about it, and I'm right, then the only reason I know to even
> tell it is because the future needs me to know and needs me to tell it
> for that future to exist.
You should advise Obama.
>
> It seems to me that such mind-numbing information would have a
> predictably large impact on a human population, which is why people
> haven't maybe really known it before; therefore, it seems likely some
> epic event is about to occur, which is allowing the information to be
> known.
if it is "mind-numbing", nobody wants to hear it or understand it.
>
> For various reasons the idea is floating around that the "end of the
> world" is in 2012. My memory is that Sir Isaac Newton actually
> calculated the correct year and he got 2010, but I've seen no mention
> of that in the record, so I'm not sure why I have that number.
go see the movie, numb-nuts.
It will tell you your past future what to tell you to do for your future
past.
>
> It may be future knowledge.
"future knowledge" = "guessing"
>
>
> James Harris
Uncle Al
>
> One of the weirder things that has emerged from my mathematical
> research
> James Harris
"Saepe errans, numquam dubitans"
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz4.htm
Owen Jacobson
> One of the weirder things that has emerged from my mathematical
> research is the possibility of continual transmission of information
> from the future to the past in order to CREATE the future, where key
> is what I call the optimal path algorithm.
>
> Used against the Traveling Salesman Problem it gives you a traveler
> going backwards in time to meet himself, where the algorithm requires
> continual communication between the two travelers in order to get the
> optimal path.
You have attached unwarranted, semi-mystical importance to the
"backwards" part of your trivial variation on bidirectional search[1].
It does not imply time travel; you can model it just as well by
thinking of two travellers, one going forward through the original
graph, and one going forward through a reversed version of the original
graph.
It'd serve you well to read up on how others have already solved graph
problems in general, and NP-complete graph problems specifically,
before running around trumpeting your "innovations". It's a pity you
won't.
-o
[1] <http://en.wikipedia.org/wiki/Bidirectional_search> for starters.
Amy
"JSH" <jst...@gmail.com> wrote in message
news:04874ca3-d17f-4100...@h40g2000prf.googlegroups.com...
<snip>
> James Harris
Man !
What a decline in intelligence their has been over the years in JSH posts.
Guess you don't read very much. Especially not math or physics.
Too bad, I used to think you were smart, but now ?
The cheese has slid off your cracker, dude.
JSH
> On 2009-11-14 23:38:49 -0500, JSH <jst...@gmail.com> said:
>
> > One of the weirder things that has emerged from my mathematical
> > research is the possibility of continual transmission of information
> > from the future to the past in order to CREATE the future, where key
> > is what I call the optimal path algorithm.
>
> > Used against the Traveling Salesman Problem it gives you a traveler
> > going backwards in time to meet himself, where the algorithm requires
> > continual communication between the two travelers in order to get the
> > optimal path.
>
> You have attached unwarranted, semi-mystical importance to the
> "backwards" part of your trivial variation on bidirectional search[1].
I checked the "bidirectional search" over a year ago when it was first
trotted out as what my optimal path algorithm does.
But the actual algorithm applied to TSP has a backwards Traveler, who
is going backwards in time to meet himself along the optimal path. In
order to calculate that path the backwards Traveler has to communicate
with himself in the past--or you can say the Traveler moving forward
in time has to communicate with himself in the future.
It's not mystical. It's an integral part of the actual algorithm,
where the optimal path once calculated collapses to a single traveler
moving forward in time.
> It does not imply time travel; you can model it just as well by
> thinking of two travellers, one going forward through the original
> graph, and one going forward through a reversed version of the original
> graph.
And you don't mention communication between the two travelers.
Now I've had over a year to ponder this thing and it scared me for a
long time, and still unsettles me, as it changes how you look at just
about everything we think we know as human beings.
But I have a simple analogy: to get, say, a college degree, when
you're a kid it can help to SEE yourself in the future getting that
degree. And from the perspective of gaining that accomplishment you
can make decisions like not to go out partying on the night before you
take your SAT's. Your imagined future self helps you make a decision
today.
What we see as conscious decision making may just be a rather basic
algorithm built into fundamental reality.
> It'd serve you well to read up on how others have already solved graph
> problems in general, and NP-complete graph problems specifically,
> before running around trumpeting your "innovations". It's a pity you
> won't.
I have read up a lot on issues in this area. But I'm also considering
an idea I have which seems to answer a lot of questions. From a
purely intellectual perspective it's a fascinating algorithm to
consider, and the implications are world shaking if true.
Then in physics optimal paths are taken by information transmission
from the future to the past and vice versa. Continually. All the
time.
So to physics information transfer backwards and forwards in time
would be just a normal way of doing business.
It's how physics even knows which way forward in time is!
> -o
>
> [1] <http://en.wikipedia.org/wiki/Bidirectional_search> for starters.
I've checked it before.
The scariest part of the optimal path algorithm I've merely hinted at
here. The scariest thing is that it may be why you think you're
conscious. Then human beings actually live in a different state
relative to time than they realize, slightly, out of phase, always
partly in the future in some sense. But to us, we're merely seeing
ourselves in some possible future.
What this algorithm says instead though is that we may be seeing
ourselves at times IN THE FUTURE, and it's not just your imagination,
but physics.
Not mysticism, but a weirder reality.
James Harris
Noob
> For various reasons the idea is floating around that the "end of the world"
> is in 2012.
sanboz
On Nov 15, 11:59 am, Owen Jacobson <angrybald...@gmail.com> wrote:
> On 2009-11-14 23:38:49 -0500, JSH <jst...@gmail.com> said:
>
> > One of the weirder things that has emerged from my mathematical
> > research is the possibility of continual transmission of information
> > from the future to the past in order to CREATE the future, where key
> > is what I call the optimal path algorithm.
>
> > Used against the Traveling Salesman Problem it gives you a traveler
> > going backwards in time to meet himself, where the algorithm requires
> > continual communication between the two travelers in order to get the
> > optimal path.
>
> You have attached unwarranted, semi-mystical importance to the
> "backwards" part of your trivial variation on bidirectional search[1].
I checked the "bidirectional search" over a year ago when it was first
trotted out as what my optimal path algorithm does.
But the actual algorithm applied to TSP has a backwards Traveler, who
is going backwards in time to meet himself along the optimal path. In
order to calculate that path the backwards Traveler has to communicate
with himself in the past--or you can say the Traveler moving forward
in time has to communicate with himself in the future.
It's not mystical. It's an integral part of the actual algorithm,
where the optimal path once calculated collapses to a single traveler
moving forward in time.
******************start message*******************
=> JOIN THE ASHTAR COMMAND
We already do this on a real time basis, get beyond primitive theory; and
free your mind of limited thoughts;
http://www.ashtarcommandcrew.net/
http://www.luisprada.com/Protected/the_ashtar_command.htm
http://ashtarcommandtruth.com/
http://www.lightvoyager.com/ACnews/accnews.html
http://ashtargalacticcommanders.ning.com/
http://www.freewebs.com/ashtargalacticcommand/
http://en.wikipedia.org/wiki/Ashtar_(extraterrestrial_being)
http://www.youtube.com/watch?v=Nu0PGd-IVNI
http://www.youtube.com/watch?v=gflfRrcn6QI
http://www.disclose.tv/action/viewvideo/6844/Ascension_Timelines__Ashtar_Command_and_2012/
***************end message*******************************
Mark Murray
> "JSH" <jst...@gmail.com> wrote in message
> news:fa5c6070-f59d-4d4a...@y10g2000prg.googlegroups.com...
...etc
Sanboz - it is very hard to see what you wrote and what you are quoting.
Please quote properly. Turning off HTML may help.
M
sanboz
"Mark Murray" <w.h....@example.com> wrote in message
news:4b019767$0$2479$db0f...@news.zen.co.uk...
it posts in text always
but it is Outlook Express -- it seems to forget the > marks on some posts
coming in, (about 1/4) although it is set to show them.
I haven't figured that one out
and on the longer posts it is a pain to go back and put in the >
Any Ideas?
Owen Jacobson
> On Nov 15, 11:59�am, Owen Jacobson <angrybald...@gmail.com> wrote:
>> On 2009-11-14 23:38:49 -0500, JSH <jst...@gmail.com> said:
>>
>>> One of the weirder things that has emerged from my mathematical
>>> research is the possibility of continual transmission of information
>>> from the future to the past in order to CREATE the future, where key
>>> is what I call the optimal path algorithm.
>>
>>> Used against the Traveling Salesman Problem it gives you a traveler
>>> going backwards in time to meet himself, where the algorithm requires
>>> continual communication between the two travelers in order to get the
>>> optimal path.
>>
>> You have attached unwarranted, semi-mystical importance to the
>> "backwards" part of your trivial variation on bidirectional search[1].
>
> I checked the "bidirectional search" over a year ago when it was first
> trotted out as what my optimal path algorithm does.
>
> But the actual algorithm applied to TSP has a backwards Traveler, who
> is going backwards in time to meet himself along the optimal path. In
> order to calculate that path the backwards Traveler has to communicate
> with himself in the past--or you can say the Traveler moving forward
> in time has to communicate with himself in the future.
>
> It's not mystical. It's an integral part of the actual algorithm,
> where the optimal path once calculated collapses to a single traveler
> moving forward in time.
There is no "time" at all as far as any NP-complete problem is
concerned, other than the number of steps needed to reach a solution.
Your algorithm proceeds stepwise through two graphs (the "forward"
graph for your "forward" traveller, which is the original complete
weighted graph, and the "backwards" graph, which is derived from the
"forward" graph by reversing the weights, for the "backwards"
traveller). At each step, your algorithm determines which node to add
to each traveller's list of nodes next based on some vaguely-specified
criteria that is a strict function of the graph and the list of nodes
visited so far. Your algorithm terminates on a step when the lists of
notes for the two travellers intersect, and the final path is the
"forward" list up to the point where the lists intersect, followed by
the reverse of the "backwards" list.
We can model your algorithm as a perfectly ordinary recursive function
-- just like any other bidirectional search. No "communication back in
time" -- the choices made at step N of your algorithm's evaluation
depend only on the choices made in steps 0..N, not on the choices made
at some step M > N. There is nowhere any requirement that an algorithm
that searches for shortest paths through a graph start at the
beginning, and starting at other points, or at multiple points is not
innovative.
If (and this is a big if) your algorithm provably produces a
Hamiltonian cycle over the input graph such that the sum of the weights
in the circuit is minimal, then your algorithm "solves" the travelling
salesman problem. If (and this is the other big if) it provably does so
in polynomial time, it is of interest to the mathematical community.
You have proven neither thing to the satisfaction of anyone besides
yourself.
>> It does not imply time travel; you can model it just as well by
>> thinking of two travellers, one going forward through the original
>> graph, and one going forward through a reversed version of the original
>> graph.
>
> And you don't mention communication between the two travelers.
The travellers are irrelevant. You could eliminate them entirely by
phrasing your algorithm in terms of adding one node to each of two
lists of nodes at each iteration without losing any details. There is
no time, there are no travellers, there is no communication, there is
only the graph, the various structures necessary to keep track of the
algorithm's state, and the repeated application of some deterministic
function to those two things.
This is math, not amateur cosmology hour.
>> It'd serve you well to read up on how others have already solved graph
>> problems in general, and NP-complete graph problems specifically,
>> before running around trumpeting your "innovations". It's a pity you
>> won't.
>
> I have read up a lot on issues in this area. But I'm also considering
> an idea I have which seems to answer a lot of questions. From a
> purely intellectual perspective it's a fascinating algorithm to
> consider
Well, yes, there are lots of ways to refine bidirectional search for
various scenarios, just like there are lots of ways to refine any other
basic graph search. There's lots of fun to be had.
> and the implications are world shaking if true.
Not even a shiver. It's been done, it's not any faster than brute force
and no known algorithm generates solutions to the travelling salesman
problem in polynomial worst-case[0] time, including various heuristic
bidirectional search algorithms. This is undergrad-level stuff, dude.
So, to reiterate, from what little you've said this time around:
- Your algorithm is not novel. It is a variation on bidirectional search.
- You have not proven that your algorithm always generates Hamiltonian
cycles over complete weighted graphs.
- You have not proven that when your algorithm does generate a
Hamiltonian cycle the cycle will have minimal weight.
- You have not proven that it does any of this in polynomial time.
All of these things must be addressed if you want to shake the world.
The rest is your own distractions.
-o
[0] or average-case time, if you insist on solving the general problem
rather than some useful subset of the problem)
Mark Murray
> Any Ideas?
Use something else? Thunderbird?
M
_...@jeff_relf.seattle.invalid
You ( sanboz ) are using an out·dated copy of Outlook Express.
To >quote lines from a Google Groups post you need to update to
Windows Live Mail ( a.k.a. Outlook Express 7 ), download it here:
Note: Passing “ /newsonly ” ( via a shortcut or a DOS prompt ),
opens wlMail in “NewsGroups-Only” mode; like this:
"C:/Program Files/Windows Live/Mail/wlMail.EXE" /newsonly
Alternatively, you could switch to ThunderBird·3, download it here:
http://download.mozilla.org/?product=thunderbird-3.0b4&os=win&lang=en-US
Rotwang
>
> [...]
>
> Now I've had over a year to ponder this thing and it scared me for a
> long time, and still unsettles me, as it changes how you look at just
> about everything we think we know as human beings.
James, I tried implementing your algorithm today, and, assuming I did
it correctly, it doesn't work. The version of the algorithm I
implemented was the one described here:
http://mymath.blogspot.com/2008/07/distance-normalized-tsp-algorithm.html
I tested my implementation of your algorithm on a number of random
graphs with n nodes, for various values of n, each of which was
generated by placing n points randomly in a two dimensional box of
side length 1, and letting the cost for travelling from one point to
another be given by the Euclidean distance between the points. The
implementation (given below) is a Python function taking a single
argument G, namely a square array of numbers (more precisely, a list
of lists, in which each element is assumed to have the same length as
the list itself), where G[i][j] represents the cost of travelling from
node i to j; nodes are labelled by integers from 0 to n - 1. An
example input for which it didn't work is:
G = [[0.0, 0.81629958572320682, 0.47104753891221535,
0.54089803432253869, 0.17625552155944968, 0.17521697257516033],
[0.81629958572320682, 0.0, 0.76217733063718074, 0.69606557418793558,
0.83552009367224145, 0.9540106420233877], [0.47104753891221535,
0.76217733063718074, 0.0, 0.10521843535006821, 0.31763962908191762,
0.44807626585131011], [0.54089803432253869, 0.69606557418793558,
0.10521843535006821, 0.0, 0.40380656862982828, 0.5408020333640231],
[0.17625552155944968, 0.83552009367224145, 0.31763962908191762,
0.40380656862982828, 0.0, 0.14300478585446333], [0.17521697257516033,
0.9540106420233877, 0.44807626585131011, 0.5408020333640231,
0.14300478585446333, 0.0]]
For this input, my implementation of your algorithm returns the path
[5,0,3,1,2,4] (start at node 5, then travel to node 0, etc., then
finally travel from 4 back to 5), with a total cost of around 2.64.
However, brute force finds the path [0,1,3,2,4,5], with a cost of
around 2.25.
Below is the Python script. It's possible that I implemented your
algorithm incorrectly, but the fact that it gave the right answer for /
most/ inputs I tried suggests that I didn't screw up too badly. The
scripts I used for generating random graphs and solving the problem by
brute force will be provided upon request, and I'm also more than
happy to answer any questions you may have about what follows.
def tsj(G):
n = range(len(G))
T = []
for i in n:
T1 = [i]
T2 = [i]
m = [j for j in n if j != i]
while len(m) > 1:
c1 = []
for j in m:
c2 = []
for k in m:
if j != k:
c2.append([k,G[T1[-1]][j]*G[T2[0]][k]])
r = 0
for q in xrange(1,len(c2)):
if c2[q][1] < c2[r][1]:
r = q
c1.append([j,c2[r][0],c2[r][1]])
r = 0
for q in xrange(1,len(c1)):
if c1[q][2] < c1[r][2]:
r = q
T1.append(c1[r][0])
T2 = [c1[r][1]] + T2
m.remove(T1[-1])
m.remove(T2[0])
S = T1 + m + T2
T.append([S,sum([G[S[j]][S[(j+1)]] for j in n])])
r = 0
for q in xrange(1,len(G)):
if T[q][1] < T[r][1]:
r = q
return [[T[r][0][i] for i in n],T[r][1]]
Jim Ferry
> "JSH" <jst...@gmail.com> wrote in message
>
> news:04874ca3-d17f-4100...@h40g2000prf.googlegroups.com...
>
> > One of the weirder things that has emerged from my mathematical
> > research is the possibility of continual transmission of information
> > from the future to the past in order to CREATE the future, where key
> > is what I call the optimal path algorithm.
>
> Join the Ashtar Command, they do that all the time. They have "channelers"
> that get furure information from Aliens in Spaceeships waiting to land, and
> beam it to Earth where it is used for profit by the Channelers.
It's not that subatomic particles are somehow performing
a computation of optimal paths: it's just that the only
place where the probability densities of events rise
above zero is at critical points, and the only critical
points are minima, as in Feynman's explanation of QED.
As James correctly observes, these transtemporal Traveling
Salesman Problems are being "solved" all the time (at a
subatomic level), although their effects are seldom
observed macroscopically, due to mass limitations.
> > For various reasons the idea is floating around that the "end of the
> > world" is in 2012. My memory is that Sir Isaac Newton actually
> > calculated the correct year and he got 2010, but I've seen no mention
> > of that in the record, so I'm not sure why I have that number.
However, once the Higgs boson is involved, macroscopic
effects are possible. The lowest energy (therefore
most probable) effects (per unit entropy) are on human
consciousness. Thus, on the one hand we have the CERN
experiment which is continually forking the universe
into branches in which (a) the universe is destroyed
(well, rendered inhospitable to life at any rate) via
an Ice-9-like disaster, and (b) some glitch delays the
disaster. It may seem amusing or fitting somehow that
the mass-imbuing Higgs boson TSP network extended back
in time to order Sir Isaac Newton's brain so as to
understand mass (as inertia) and leave the correct
date (2010) for its own creation / world's destruction,
but the world's a funny old place.
Anyway, James has warned you. I've embellished the
warning. Now what are you going to do? Nothing?
Yeah, me neither. I feel surprisingly relaxed about
the world ending next year. Maybe that's because I'm
not insane. I'm only bringing you this message of
doom because my brain is somehow being entangled in
the Higgs boson TSP via which the future is
communicating with James Harris and Isaac Newton.
My brain and my even-now-typing fingers have no
choice in this matter. But to believe it? That
would be crazy indeed!
A
James' initial warning was the first herald of the coming wave of
Trans-temporal Travelling Salesman (TTS) Consciousness. While it was
his (future-projected) genius (travelling back to his past-present
self) that first recognized the coming wave of TTS Consciousness, it
seems to be spreading rapidly--first you were "infected," and now
today I have suddenly come to understand as well that James' solution
to the Travelling Salesman Problem has opened up paths of
communication between the future universe and the present universe
which are large enough in scale that my aware-of-TTS-Consciousness
future self has informed my present-day self of the coming of TTS
Consciousness.
Since James' solution to the Travelling Salesman Problem will seem
obvious to the world as TTS Consciousness becomes widespread, don't
you think it is our duty to help announce to the world that it was
James who first discovered the solution to the Travelling Salesman
Problem, so that his name is properly recognized in history? After
all, while we now know that the future sends information back to the
past, we know very well (from history) that the past doesn't always
send all the right information forward to the future.
Jim Ferry
Is it our duty? Perhaps. I'm finding it difficult to adapt such
concepts as duty from my previous mental framework, in which I was
a being with free will, to this new TTS Consciousness. Whether I will
be greeting the dawn of this Harristotelian age by shouting the glory
of James from the rooftops is a matter for the future enlightened,
transcendent-genius-Harris to decide for me.
sanboz
I'll check it out with some others, see what happens
Aleph
_@Jeff_Relf.Seattle.inValid says...
>
> ?
> You ( sanboz ) are using an outï¿œdated copy of Outlook Express.
>
> To >quote lines from a Google Groups post you need to update to
> Windows Live Mail ( a.k.a. Outlook Express 7 ), download it here:
Who died and made you the king of what news clients people use?
Peter
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
Bill Snyder
<Usene...@gishpuppy.com> wrote:
>In article <_@Jeff_Relf.Seattle.2009_Nov17.0.00am.Qp>,
>_@Jeff_Relf.Seattle.inValid says...
>>
>> ?
>> You ( sanboz ) are using an out·dated copy of Outlook Express.
>>
>> To >quote lines from a Google Groups post you need to update to
>> Windows Live Mail ( a.k.a. Outlook Express 7 ), download it here:
>
>Who died and made you the king of what news clients people use?
His next-to-last brain cell.
--
Bill Snyder [This space unintentionally left blank]
sanboz
<_@Jeff_Relf.Seattle.inValid> wrote in message
news:_@Jeff_Relf.Seattle.2009_Nov17.0.00am.Qp...
Ill check them both out, TBird is probably better
Aleph
bsn...@airmail.net says...
>
> On Tue, 17 Nov 2009 23:00:50 -0000, Aleph
> <Usene...@gishpuppy.com> wrote:
>
> >In article <_@Jeff_Relf.Seattle.2009_Nov17.0.00am.Qp>,
> >_@Jeff_Relf.Seattle.inValid says...
> >>
> >> ?
> >>
> >> To >quote lines from a Google Groups post you need to update to
> >> Windows Live Mail ( a.k.a. Outlook Express 7 ), download it here:
> >
> >Who died and made you the king of what news clients people use?
>
> His next-to-last brain cell.
I pity his last brain cell. I suspect it has also gone and now Jeff is
simply acting as an automaton....
JSH
> On 16 Nov, 03:32, JSH <jst...@gmail.com> wrote:
>
>
>
> > [...]
>
> > Now I've had over a year to ponder this thing and it scared me for a
> > long time, and still unsettles me, as it changes how you look at just
> > about everything we think we know as human beings.
>
> James, I tried implementing your algorithm today, and, assuming I did
> it correctly, it doesn't work. The version of the algorithm I
> implemented was the one described here:
>
Yup. Unfortunately.
You appear to have skipped one step though maybe I missed it in your
code.
The final "cost" is the distance from the respective nodes, so, say,
you consider a move from node 0 for T_1 to node 1 and for T_2 to node
2, then you need the distance from node 1 to node 2 multiplied times
the cost*distance from 0 to node 1 for T_1, and the distance from node
1 to node 2 multiplied times the cost*distance for T_2.
The algorithm seeks to minimize the distance that T_1 and T_2 ever get
from each other, so they stay as close as possible to each other which
means that the distance from the nodes to which they move is part of
the equation.
If that is in your code I missed it. 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? If so, and you
DID leave out a key piece that could be of research interest as to
why.
Thanks for the implementation!
I normally delete out code in reply as it's so messy displayed in
posts but Python displays ok so leaving it.
Looking over the code, I think I have the gist of it, and I'm not sure
I see where the distance from the prospective nodes to which to move
to each other is part of the calculation. But I could have just
missed that.
Presumably here's key code:
<quote>
> if j != k:
> c2.append([k,G[T1[-1]][j]*G[T2[0]][k]])
> r = 0
> for q in xrange(1,len(c2)):
> if c2[q][1] < c2[r][1]:
> r = q
> c1.append([j,c2[r][0],c2[r][1]])
</quote>
Looks to me like a straight original position to possible node
calculation for "T1" and "T2" where you add the minimum found to "c1"
and "c2", but I don't see where the distance from the potential node
position for "T1" is multiplied times the one for "T2".
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.
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.
Thanks!!!
James Harris
Rotwang
> On Nov 17, 1:52 am, Rotwang <sg...@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:
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
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
per 1000 attempts with 6 nodes, and about 440 failures per 1000
attempts with 10 nodes). Note that the random graphs I've been using
as input are all of a very specific type, though I also tried
generating graphs by sticking points at random in boxes of dimensions
other than 2, and it didn't seem to make much difference.
> If so, and you
> DID leave out a key piece that could be of research interest as to
> why.
I don't think this is as surprising as you seem to believe. I expect
(but haven't checked) that a dumb-as-rocks greedy algorithm that
simply sent the traveller to the nearest node not yet visited would
also give the right answer for most inputs I tried; it's just the
nature of the problem (much like factoring) that most inputs are easy
to solve.
> Thanks for the implementation!
My pleasure, I had some time to kill.
Yes. At the start of each iteration of the "while" loop, T1 and T2 are
lists containing the paths taken so far by the forward traveller and
backwards traveller (the latter in reverse order, since it makes what
follows more compact), and m is a list of nodes that have not yet been
visited. T1[-1] returns the last element of T1, and T2[0] returns the
first element of T2. j and k are the moves that the forward and
backwards travellers are considering, and c2 is a list of all possible
values of cost*cost as k varies with j fixed; at the end of each "for
k in ..." loop, the program finds the smallest element of c2 and
appends it to c1, then when all possible values of j have been tried,
the program checks for the smallest element of c1 and adds the
corresponding values of j and k to T1 and T2, and also removes them
from m.
> but I don't see where the distance from the potential node
> position for "T1" is multiplied times the one for "T2".
>
> Good work though with what you have. Can you give an example where it
> WORKS as well.
Sure, here's one with 6 nodes:
[[0.0, 0.17448009086755145, 0.55623052938910778, 0.10128874519356396,
0.49952861243264768, 0.40610612562053522], [0.17448009086755145, 0.0,
0.39057755479918704, 0.23512859628257424, 0.59434207506913084,
0.38352865389624907], [0.55623052938910778, 0.39057755479918704, 0.0,
0.58098483474237039, 0.80736343868080684, 0.66137691609161453],
[0.10128874519356396, 0.23512859628257424, 0.58098483474237039, 0.0,
0.39824449203958673, 0.50715021950156514], [0.49952861243264768,
0.59434207506913084, 0.80736343868080684, 0.39824449203958673, 0.0,
0.90480204601747161], [0.40610612562053522, 0.38352865389624907,
0.66137691609161453, 0.50715021950156514, 0.90480204601747161, 0.0]]
Both my implementation of your algorithm and brute force give the path
[5,0,3,4,2,1].
Mark Murray
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
JSH
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.
> per 1000 attempts with 6 nodes, and about 440 failures per 1000
> attempts with 10 nodes). Note that the random graphs I've been using
But by here that is insanely too good.
Now you have 10! = 3628800. With 10*10 you get 100, or a really small
success rate of 100/3628800.
> as input are all of a very specific type, though I also tried
> generating graphs by sticking points at random in boxes of dimensions
> other than 2, and it didn't seem to make much difference.
>
> > If so, and you
> > DID leave out a key piece that could be of research interest as to
> > why.
>
> I don't think this is as surprising as you seem to believe. I expect
> (but haven't checked) that a dumb-as-rocks greedy algorithm that
> simply sent the traveller to the nearest node not yet visited would
> also give the right answer for most inputs I tried; it's just the
> nature of the problem (much like factoring) that most inputs are easy
> to solve.
Greedy algorithm is easier to implement. You just pick the closest
node.
The success rate you're giving above though is too crazy good, so I
doubt the greedy algorithm could match it.
Of course I could be mis-reading what you said above about success, as
that is an insanely good success rate despite not being perfect.
Way too good. I suspect you have a mistake or I'm reading you wrong.
By 10 nodes you have 3628800 potential paths. Over 3 million.
So it should drop by 2 with each iteration in the most inner loop.
> first element of T2. j and k are the moves that the forward and
> backwards travellers are considering, and c2 is a list of all possible
> values of cost*cost as k varies with j fixed; at the end of each "for
> k in ..." loop, the program finds the smallest element of c2 and
> appends it to c1, then when all possible values of j have been tried,
> the program checks for the smallest element of c1 and adds the
> corresponding values of j and k to T1 and T2, and also removes them
> from m.
Interesting. And I guess in there somewhere is the rule to just
append the remaining node if only one node is left.
That only occurs with an even number of nodes.
So to which do you append? Your "T1" or your "T2"?
It doesn't matter. I think for the algorithm I just picked T_1.
> > Good work though with what you have. Can you give an example where it
> > WORKS as well.
>
> Sure, here's one with 6 nodes:
>
> [[0.0, 0.17448009086755145, 0.55623052938910778, 0.10128874519356396,
> 0.49952861243264768, 0.40610612562053522], [0.17448009086755145, 0.0,
> 0.39057755479918704, 0.23512859628257424, 0.59434207506913084,
> 0.38352865389624907], ...
>
> read more »
Interesting. The algorithm will do 6*6 = 36 checks when there are 6!
= 720 possible paths.
In any event, interesting data from the effort. Hope you screwed up
as the algorithm sounds rather good otherwise.
My hope is that it'd just be crap. Intermediate where it does as well
as you claim may mean you made a mistake or there is a fix that will
make it perfect.
That would really be bad as I'm not interested in figuring out any
more fixes. I'm cruising now and not interested in doing more
original research.
So then! Explain to me how I mis-interpreted your too good to be true
results above.
James Harris
spudnik
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
Rotwang
On 28 Nov, 22:58, JSH <jst...@gmail.com> wrote:
> On Nov 28, 10:55 am, Rotwang <sg...@hotmail.co.uk> wrote:
>
> [...]
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
is invariant under cyclic permutations and order reversals, so there
are 2*m paths which all have the shortest total cost. Secondly, your
algorithm is O(m^3), not O(m^2) - not that I see the relevance, since
the algorithm doesn't simply try m^3 guesses at random so there is no
earthly reason to expect that it should be correct only 2*m^3/(m - 1)!
of the time.
> > per 1000 attempts with 6 nodes, and about 440 failures per 1000
> > attempts with 10 nodes). Note that the random graphs I've been using
>
> But by here that is insanely too good.
>
> Now you have 10! = 3628800. With 10*10 you get 100, or a really small
> success rate of 100/3628800.
Nobody has ever claimed that your algorithm works by trying m^2 paths
at random, so I fail to see the import of the fact that it performs
better than trying m^2 paths at random would.
> > as input are all of a very specific type, though I also tried
> > generating graphs by sticking points at random in boxes of dimensions
> > other than 2, and it didn't seem to make much difference.
>
> > > If so, and you
> > > DID leave out a key piece that could be of research interest as to
> > > why.
>
> > I don't think this is as surprising as you seem to believe. I expect
> > (but haven't checked) that a dumb-as-rocks greedy algorithm that
> > simply sent the traveller to the nearest node not yet visited would
> > also give the right answer for most inputs I tried; it's just the
> > nature of the problem (much like factoring) that most inputs are easy
> > to solve.
>
> Greedy algorithm is easier to implement.
A bit, yes. See below for my implementation.
> You just pick the closest
> node.
>
> The success rate you're giving above though is too crazy good, so I
> doubt the greedy algorithm could match it.
I tried coding a version of the greedy algorithm I mentioned earlier,
to see how it compared. Just now, when I tried running both your
algorithm and the greedy algorithm on 1000 random graphs with 10
nodes, your algorithm gave the wrong answer 510 times, while the
greedy algorithm gave the wrong answer 548 times. Yours does better,
but not that much better, and the greedy algorithm (itself O(m^3))
also has a much higher success rate than your m^2/m! guess.
A couple of comments - although I said before that your algorithm gave
about 440 failures per 1000 inputs with m = 10, this was from memory,
and it's possible I got it wrong - the discrepancy between 440 and 510
does seem rather too large to be a purely statistical fluctuation.
Perhaps the 440 was with m = 9. Also, the difference in success rate
between your algorithm and the greedy one looked more pronounced when
I tried running both on graphs with fewer nodes.
In the time I've been writing this reply I've run the same test again.
This time your algorithm gave the wrong answer 473 times, while the
greedy one gave the wrong answer 565 times. So I guess you got unlucky
first time around, but my above conclusions still stand.
> Of course I could be mis-reading what you said above about success, as
> that is an insanely good success rate despite not being perfect.
Not really, compared to that of the greedy algorithm.
> Way too good. I suspect you have a mistake or I'm reading you wrong.
>
> By 10 nodes you have 3628800 potential paths. Over 3 million.
>
> [...]
>
> > > Looks to me like a straight original position to possible node
> > > calculation for "T1" and "T2" where you add the minimum found to "c1"
> > > and "c2",
>
> > Yes. At the start of each iteration of the "while" loop, T1 and T2 are
> > lists containing the paths taken so far by the forward traveller and
> > backwards traveller (the latter in reverse order, since it makes what
> > follows more compact), and m is a list of nodes that have not yet been
> > visited. T1[-1] returns the last element of T1, and T2[0] returns the
>
> So it should drop by 2 with each iteration in the most inner loop.
Not in the most inner loop (which loops over possible values of k),
but with each iteration of the "while" loop.
> > first element of T2. j and k are the moves that the forward and
> > backwards travellers are considering, and c2 is a list of all possible
> > values of cost*cost as k varies with j fixed; at the end of each "for
> > k in ..." loop, the program finds the smallest element of c2 and
> > appends it to c1, then when all possible values of j have been tried,
> > the program checks for the smallest element of c1 and adds the
> > corresponding values of j and k to T1 and T2, and also removes them
> > from m.
>
> Interesting. And I guess in there somewhere is the rule to just
> append the remaining node if only one node is left.
Yes, in the line
S = T1 + m + T2
which runs after the "while" loop has finished, i.e. when m is either
empty or has a single element. Adding lists together in Python
concatenates them, so e.g. if T1 = [0,1,2], T2 = [4,5,0] and m = [3]
then S = [0,1,2,3,4,5,0] (this is the reason why successive moves of
the backwards traveller are added to the beginning of T2, rather than
the end - it means that the above definition of S automatically gives
the collapsed path in the correct order). If m is empty then the "+ m"
does nothing, obviously.
> That only occurs with an even number of nodes.
>
> So to which do you append? Your "T1" or your "T2"?
>
> It doesn't matter. I think for the algorithm I just picked T_1.
Indeed, unless I have completely misunderstood your description of the
algorithm it obviously makes no difference, since you concatenate the
paths T1 and T2 to make the collapsed path anyway.
> [...]
>
> My hope is that it'd just be crap. Intermediate where it does as well
> as you claim may mean you made a mistake or there is a fix that will
> make it perfect.
I doubt it. Don't know much about the problem myself, but I would
guess (and this guess is apparently confirmed by the results of my
various tests) that it's easy enough to come up with a crude algorithm
that works a lot of the time, and that various refinements will
marginally improve the results. But I don't see how any such
refinements will be able to handle all possible pathologies that occur
(I suppose that would require some fairly radical new ideas, or else
someone would have already solved the problem).
> So then! Explain to me how I mis-interpreted your too good to be true
> results above.
See above - I don't believe that "too good to be true" is anything
more than wishful thinking.
Here's the greedy algorithm script:
def tsg(G):
n = range(len(G))
T = []
for i in n:
T1 = [i]
m = [j for j in n if j != i]
while len(m) > 0:
r = 0
for q in xrange(1,len(m)):
if G[T1[-1]][m[q]] < G[T1[-1]][m[r]]:
r = q
T1.append(m[r])
m.remove(T1[-1])
T.append((T1,sum([G[T1[j]][T1[(j+1)%len(n)]] for j in n])))
r = 0
for q in xrange(1,len(n)):
JSH
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
JSH
> One of the weirder things that has emerged from my mathematical
> research is the possibility of continual transmission of information
> from the future to the past in order to CREATE the future, where key
> is what I call the optimal path algorithm.
Which it looks like the poster "Rotwang" has debunked as a solution!
So all my wild speculations in this post go out the window.
> Used against the Traveling Salesman Problem it gives you a traveler
> going backwards in time to meet himself, where the algorithm requires
> continual communication between the two travelers in order to get the
> optimal path.
Which it seems does not work, according to the work done by "Rotwang"
which is in this thread.
> If that is a routine part of nature then light takes the optimal path
> in that way, and it also gives an arrow to time--we think we're
> traveling forward in the future as we're the collapsed path, when
> actually we're traveling both forwards AND backwards in time.
>
> The collapse to an optimal path gives us the illusion of only going
> forwards in time.
>
> The arrow is the collapsed optimal path which appears to only go
> forward in time.
>
> If so, then some of our "history" can be information transmitted to
> the past in order to create our future (and our present).
Trashing that idea for the realworld, but I might still use it someday
for science fiction.
So, yup, nothing to see here, move along. I was just wrong.
Happens all the time.
James Harris
Mark Murray
> So I didn't solve the TSP problem and did not prove P=NP. Oh well.
> Not a big deal.
Apologies for the insults left along the way?
M
Mark Murray
> So, yup, nothing to see here, move along. I was just wrong.
>
> Happens all the time.
Noted.
Now look at your incorrect SWJPAM paper.
M
Peter
> So I didn't solve the TSP problem and did not prove P=NP. Oh well.
> Not a big deal.
>
> James Harris
Such a nonchalant attiutude for a person who holds the fate of the
entire world in his hands (giggle), although maybe, just maybe, after
this your 50th (or is it the 500th) unsuccessful attempt, one might be
tempted to side with dear Uncle Al, who so poetically put it to you in
his classic "An Ode to James Harris"
"Somebody said it couldn't be done,
But James with a chuckle replied
That "maybe it couldn't," but he would be one
Who wouldn't say so till he'd tried.
So James buckled right in with the trace of a grin
On his face. If he worried he hid it.
James started to sing and he tackled the thing
And James never fucking could do it.
Somebody scoffed: "Oh, you'll never do that;
At least no one has ever done it";
But James took off his coat and he took off his hat,
And the first thing we knew he'd begun it.
With a lift of his chin and a bit of a grin,
Without any doubting or quiddit,
James started to sing and he tackled the thing
And James never fucking could do it.
There are thousands to tell James it cannot be done,
There are thousands to prophesy failure;
There are thousands to point out to James, one by one,
"How hopeless that task set before you."
But just buckle in with a bit of a grin,
James take off your coat and go to it;
Just start to sing as you tackle the thing
And James, you'll never fucking do it. "
HTH,
Peter
Peter
Isn't the comedy itself sufficient?
Peter
Mark Murray
> Isn't the comedy itself sufficient?
Naah. Like James, I like dreaming.
M
JSH
> On Nov 29, 10:22 am, JSH <jst...@gmail.com> wrote:
>
> > So I didn't solve the TSP problem and did not prove P=NP. Oh well.
> > Not a big deal.
>
> > James Harris
>
> Such a nonchalant attiutude for a person who holds the fate of the
> entire world in his hands (giggle), although maybe, just maybe, after
Well, pity me for daring to acknowledge error!
And people wonder why it's so hard to do it?
Because nasty people will insult you for doing it, that's why.
These people simply hate. Making yourself vulnerable by acknowledging
error is an invitation to them to kick you while you are down. They
are dismal humans.
> this your 50th (or is it the 500th) unsuccessful attempt, one might be
> tempted to side with dear Uncle Al, who so poetically put it to you in
You mean the robot? What makes you think his behavior is not
predictable?
> his classic "An Ode to James Harris"
<deleted>
I've read it before.
James Harris
Mark Murray
> Well, pity me for daring to acknowledge error!
You? Challenge error?
HAH!
JSH: "There is no point looking for errors there since there
aren't any" (paraphrased).
> And people wonder why it's so hard to do it?
I Just wonder why its so hard for you to generalise the experience.
You finally admitted that your TSP solution was an abject failure
(never mind the "fact" that it was a proof). Now keep going.
> Because nasty people will insult you for doing it, that's why.
What about your own insults (the ones you haven't dealt with yet,
even though your victims have been long vindicated).
> These people simply hate. Making yourself vulnerable by acknowledging
> error is an invitation to them to kick you while you are down. They
> are dismal humans.
Look in the mirror.
>> this your 50th (or is it the 500th) unsuccessful attempt, one might be
>> tempted to side with dear Uncle Al, who so poetically put it to you in
>
> You mean the robot? What makes you think his behavior is not
> predictable?
"The Robot" was right. You were wrong. Deal with it.
> I've read it before.
And you will do so again. And again &c, until you learn.
M