The conventional halting problem

[wij]

This is a copy from https://groups.google.com/g/comp.lang.c++/c/PbhSSy66We8
Title: Re: New test program (Succeeds V2)[ Honest Dialogue? ]( maybe not )
=========
On 5/8/2021 10:55 PM, wij wrote:
>>> ...
>>> The non-halting criteria of (1)(2)(3)(4) mentions X(),Y(), while the examples
>>> use u32 Halts(u32,u32), and H_Hat(u32). I am baffled what is really said.
>>> (consistency)
>>>
>> The X() and Y() function names specify the generic [infinite-recursion]
>> behavior pattern. This behavior pattern equally applies to the
>> [infinitely nested simulation] behavior of H_Hat() relative to Halts().
>>
>> I cut-and-pasted this stuff from my five page paper. The above is
>> explained on page one of my paper.
>>
>> http://www.liarparadox.org/Halting_problem_undecidability_and_infinite_recursion.pdf
>> --
>
> How is the result of your paper helpful in practice (or in theory)?
>

The conventional halting problem proofs cease to prove that the halting
problem cannot be solved. This provides the basis for the solution to
the halting problem. https://en.wikipedia.org/wiki/Halting_problem

The fundmental limits of algorithmic computation that were previously
demonstrated by the halting problem proofs cease to exist.
==========

And, an article: The Halting Paradox
https://arxiv.org/pdf/1906.05340.pdf


I think "The conventional halting problem" is easy to understand and
reproducible. The above two dissenting views are not so understandable
and reproducible if not misleading.
Put aside those abstract, theoretical, technical models and terms.

Let TEST_HALT be a program: TEST_HALT <pgm>
TEST_HALT <pgm> will exit and the exit status is 0 if <pgm> halt, 1 otherwise.

Can such a program TEST_HALT can ever be correctly written?
And, what does "The fundamental limits of algorithmic computation that
were previously demonstrated by the halting problem proofs cease to
exist." mean? We are still program (your program, too) in the TM realm, aren't we?

[olcott]

The key element of all the conventional halting problem proofs is that
their counter-example specifies decidable infinitely nested simulation
to every halt decider that bases it halting decision on simulating its
input. This is not the same thing that the other author is saying.

--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre
minds." Einstein

[Richard Damon]

It is only 'decidably infinite' by your fault logic. When you include
the fact that Halts needs to be a Halting Computation in the processing
of the trace, this logic is shown to be incorrect.

[wij]

Your paper, like others I saw, is composed of many irrelevant/impovised sub-theories.
Each of them needs lots more to backup its correctness., thus, incomplete by itself.
Worst, those sub-theories have little causal conjunctions.
Basically, the quality of your paper is of very low value.

Simple test:
Build a program: TEST_HALT <pgm> // both TEST_HALT, <pgm> are executable programs

TEST_HALT <pgm> will exit and the exit status is 0 if <pgm> halt, 1 otherwise.

If you can publish TEST_HALT, everything would be clearer. Otherwise just smooth talk.
I think the latter is the truth. You can not write TEST_HALT but follow up with more
logic or you own (smooth talk).

[olcott]

**My whole proof is boiled down to as simple as this:**
When the execution trace of H_Hat() correctly matches a correct
non-halting behavior pattern then the decision of non-halting is
impossibly incorrect: AKA modal logic necessarily(P)

When an execution trace matches the following criteria we know that the
non-halting behavior of infinite recursion or infinitely nested
simulation has been matched:

If the execution trace of function Y() shows:
(1) function X() is called twice in sequence from the same machine
address of Y()
(2) with the same parameters to X()
(3) with no conditional branch or indexed jump instructions in Y()
(4) with no function call returns from X()
then the function call from Y() to X() is infinitely recursive.

The execution trace of H_Hat() shows that its behavior matches the above
non-halting criteria thus the only possible way that Halts(H_Hat, H_Hat)
== 0 is incorrect is that the criteria is incorrect.

http://www.liarparadox.org/Halting_problem_undecidability_and_infinite_recursion.pdf

[wij]

On Monday, 10 May 2021 at 20:57:26 UTC+8, olcott wrote:
>...

> **My whole proof is boiled down to as simple as this:**
> When the execution trace of H_Hat() correctly matches a correct
> non-halting behavior pattern then the decision of non-halting is
> impossibly incorrect: AKA modal logic necessarily(P)
>
> When an execution trace matches the following criteria we know that the
> non-halting behavior of infinite recursion or infinitely nested
> simulation has been matched:
>
> If the execution trace of function Y() shows:
> (1) function X() is called twice in sequence from the same machine
> address of Y()
> (2) with the same parameters to X()
> (3) with no conditional branch or indexed jump instructions in Y()
> (4) with no function call returns from X()
> then the function call from Y() to X() is infinitely recursive.
>
> The execution trace of H_Hat() shows that its behavior matches the above
> non-halting criteria thus the only possible way that Halts(H_Hat, H_Hat)
> == 0 is incorrect is that the criteria is incorrect.
>
> http://www.liarparadox.org/Halting_problem_undecidability_and_infinite_recursion.pdf
> --
> Copyright 2021 Pete Olcott
>
> "Great spirits have always encountered violent opposition from mediocre
> minds." Einstein

I have to skip all (except really compelling) your excuses, because I saw too
many and you probably also get tired to repeat them.

Nowadays, as more and more people RECEIVE messages passively, unconsciously or
'forcibly' (The book I am reading refers such people as 'sophomores'), the real knowledge
is just 'words'. The persons involved will never know they are 'sophomores'. There is a little
chance I might be one.
Sophomore (Greek) https://en.wikipedia.org/wiki/Sophomore

So the question is simple:

Build a program: TEST_HALT <pgm> // both TEST_HALT, <pgm> are executable programs

TEST_HALT <pgm> will exit and the exit status is 0 iff <pgm> halt, 1 otherwise.

It does not matter if TEST_HALT takes long time to complete. The key point is
reproducibility. So far, no one (except you?) can build TEST_HALT from anything you posted.

[olcott]

I am not sure but it seems that you are making the same mistake as Mr
Flibble in conflating the refutation of the conventional halting problem
proofs with solving the halting problem. I am doing the former and not
the latter. The former is not the same as the latter.

[olcott]

Halts() does correctly decide not halting on H_Hat() and both Halts()
and H_Hat() are correct x86 machine code that does execute.

[Jeff Barnett]

Let's include a quote from Wikipedia to clarify what you are saying
about our favorite typing demon, good old Pete.

A sophomoric view is one held confidently because of lack of awareness
of one's own ignorance. From the Greek words σοφός
(transliterated as
sophós), meaning wise or clever, and μωρός (transliterated as mōrós)
meaning foolish or stupid, related to the word moron.

> So the question is simple:
> Build a program: TEST_HALT <pgm> // both TEST_HALT, <pgm> are executable programs
> TEST_HALT <pgm> will exit and the exit status is 0 iff <pgm> halt, 1 otherwise.
>
> It does not matter if TEST_HALT takes long time to complete. The key point is
> reproducibility. So far, no one (except you?) can build TEST_HALT from anything you posted.

--
Jeff Barnett

[wij]

I was just afraid of his reply might be. Given a time period, the same drama might reoccur.
Mimic the way Chris might do.
https://www.youtube.com/watch?v=4QBl-BuEIV8

If someone can identify the symptom (a kind of brain damage to me), tell him, please.

[dklei...@gmail.com]

Peter Olcott is somewhere on the Asperger's spectrum.

He already knows that and, if I remember correctly, has admitted as much.

[olcott]

I retested myself. I am currently below the average on that scale.

[olcott]

On 5/10/2021 6:28 PM, dklei...@gmail.com wrote:

Apparently Turing himself was an Aspy.

[wij]

I got 20 points from the online test:
https://www.clinical-partners.co.uk/for-adults/autism-and-aspergers/adult-autism-test/adult-autism-test-results/results

These results indicate that...
you have a strong likelihood of Asperger’s Syndrome or autism.
--------

[olcott]

Some of the most important people in the world are Aspies.
Apparently it increases analytical skill at the expense of people skill.

[olcott]

I got an 8, one point above none at all.

[wij]

I was trying to find a name suitable to the symptom or phenomenon I discerned
from you posts. "Asperger" is apparently not the right one.
I can not say specifically what the symptom or phenomenon is but a guess some
part of the brain is physically impaired (e.g. by a tumor), which blocked your
(not mine) comprehension, so I need not delve into your theory, arguing is futile.

[olcott]

The alternative is that if you really paid attention and had sufficient
technical competence you would be able to see that I am correct.

If you cannot understand that the following sentence <is> a truism, that
would be sufficient evidence that you do not have sufficient technical
competence.

Truism:
Every simulation that would never stop unless Halts() stops it at some
point specifies infinite execution.

The following paper provides the context for the above statement.

[Kaz Kylheku]

On 2021-05-11, olcott <No...@NoWhere.com> wrote:
> The alternative is that if you really paid attention and had sufficient
> technical competence you would be able to see that I am correct.

How can you pronounce yourself correct while you have not yet produced
the corrected criteria for detecting indefinite simulation/recursion?

You've just been shown incorrect.

And, at no time have you shown yourself to be correct.

All you have is unproven conjectures.

A person proposing a conjecture is /never, ever/ regarded as correct.
The conjecture must be proven.

Your conjectures contradict established proofs.

If a conjecture contradicts an established proof, we simply refer
to the proof; the proof shows the conjecture to be invalid.

The conjecture can only be true if the proof has a mistake. (And even
then, the conjecture still needs a separate proof its own correctness.)

You have no proof of your conjecture, and you are not able to point
out where any established proofs are going wrong.

You're not qualified to even gengage the content of those proofs on
their level.

[Jeff Barnett]

So that definitely eliminates that affliction for you since you have
neither skill. Congratulation!
--
Jeff Barnett

[wij]

Luring me into your infinite recursion trap? How honest you are!
You need to get out of you own trap (or cave).
https://en.wikipedia.org/wiki/Novum_Organum#The_Idols_(Idola)

[Chris M. Thomasson]

https://youtu.be/t_FRWUPcR7Y?t=38

lol.