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computers / comp.ai.philosophy / Re: Refuting the Peter Linz Halting Problem Proof V6 [ honest dialogue is mandated ]

SubjectAuthor
* Refuting the Peter Linz Halting Problem Proof V6olcott
+* Re: Refuting the Peter Linz Halting Problem Proof V6olcott
|+* Re: Refuting the Peter Linz Halting Problem Proof V6 [ behavior ofolcott
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| `- Re: Refuting the Peter Linz Halting Problem Proof V6olcott
+* Re: Refuting the Peter Linz Halting Problem Proof V6 [ infiniteolcott
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|`- Re: Refuting the Peter Linz Halting Problem Proof V6 [ strawman errorolcott
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 | |      |       |           `* Re: Refuting the Peter Linz Halting Problem Proof V6 [ honestolcott
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 | |      `- Re: Refuting the Peter Linz Halting Problem Proof V6 [ strawman errorolcott
 | `- Re: Refuting the Peter Linz Halting Problem Proof V6 [ strawman error ]Don Stockbauer
 `- Re: Refuting the Peter Linz Halting Problem Proof V6 [ strawman errorolcott

Pages:12345
Subject: Re: Refuting the Peter Linz Halting Problem Proof V6 [ honest dialogue is mandated ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sun, 3 Apr 2022 03:12 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22
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Subject: Re: Refuting the Peter Linz Halting Problem Proof V6 [ honest
dialogue is mandated ]
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Followup-To: comp.theory
From: NoO...@NoWhere.com (olcott)
In-Reply-To: <874k3bf37l.fsf@bsb.me.uk>
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On 4/2/2022 6:57 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/2/2022 6:07 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/1/2022 7:38 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/1/2022 1:29 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/1/2022 11:38 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

I will not discuss your question until after we have mutual agreement
on all these points because I will not tolerate a one way dialogue
that only has denigration as feedback.
It's two-way.  I ask a simple question: please complete the following
line for me:

       H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊦* H.q?

and you reply without answering it.  Why don't you say you don't know
the answer to this question as well?  That would complete your admission
of knowing almost nothing about the halting problem.

(It's three times unanswered now.)

The following conclusively proves that embedded_H correctly determines
the halt status of its input.

Four times unanswered.

Unless and until we have mutual agreement on these key points
(they are the fruition of 17 years worth of work)
we cannot move on to your question.

Nothing is stopping you from answering the question.  I'm including "I
don't know" as a possible answer of course.

MUTUAL AGREEMENT ON THIS IS MANDATORY FOR FURTHER DIALOGUE

No it isn't.  Not in the passive voice.  /You/ may choose not address
these questions

Refusing to address questions stops the dialogue,
The "di" in dialogue means two.
The two of us are exchanging views.  You are making unreasonable
demands, and I am asking simple questions that you won't answer.  This
is a very informative dialog since it follows the pattern of all crank
threads.

MUTUAL AGREEMENT ON THIS IS MANDATORY FOR FURTHER DIALOGUE
Here's where we are.  After 17 years work on this you can not even say
what string must be passed to H so that H can tell us whether or not Ĥ
applied to ⟨Ĥ⟩ halts, and you won't say what state H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩
transitions to even though everyone is happy to accept that
    Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
from which the answer is easily obtained.  It would be extraordinary but
we know what's going on...

I did figure out the correct analysis of that.
The problem is that the result of the analysis is so counter-intuitive
that you will never pay enough attention to see that it is correct.

Good excuse not to say what string must be passed to H so that H can
tell us whether or not Ĥ applied to ⟨Ĥ⟩ halts and what state H.q0 ⟨Ĥ⟩
⟨Ĥ⟩ transitions to!

Keep not answering.  What else can you do?

It is very very good and helpful that you brought this question up.
It is a key issue that must be addressed. I have now fully addressed this issue. I really need to start locking in my progress though.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re:_Refuting_the_Peter_Linz_Halting_Problem_Proof_V6_[_honest_dialogue_is_mandated_]_[_Full_reply_to_André_]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sun, 3 Apr 2022 15:49 UTC
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Subject: Re:_Refuting_the_Peter_Linz_Halting_Problem_Proof_V6_
[_honest_dialogue_is_mandated_]_[_Full_reply_to_André_
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From: NoO...@NoWhere.com (olcott)
Followup-To: comp.theory
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On 4/3/2022 6:26 AM, Richard Damon wrote:

On 4/2/22 11:09 PM, olcott wrote:
On 4/2/2022 9:34 PM, Richard Damon wrote:
On 4/2/22 9:53 PM, olcott wrote:
On 4/2/2022 8:31 PM, Richard Damon wrote:
On 4/2/22 9:20 PM, olcott wrote:
On 4/2/2022 8:17 PM, Richard Damon wrote:
On 4/2/22 9:06 PM, olcott wrote:
On 4/2/2022 7:48 PM, Richard Damon wrote:
On 4/2/22 8:36 PM, olcott wrote:
On 4/2/2022 7:23 PM, André G. Isaak wrote:
On 2022-04-02 18:12, olcott wrote:
On 4/2/2022 6:58 PM, André G. Isaak wrote:

Yes, if you abort the simulation, the simulation does not reach a final state, but it is obvious that if your 'SHD' had not aborted the simulation, it *would* have ultimately reached a final state.

So an infinite loop eventually halts if you just wait long enough.

Of course not. But there is no such loop involved in the case under consideration. This is obvious to everyone participating in this thread other than you.

André


Even Richard knows that this never stops running unless aborted:

When Ĥ is applied to ⟨Ĥ⟩
   Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then embedded_H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

Then these steps would keep repeating:
   Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩
   Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
   Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H2 simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩...



But the loop isn't an infinite 'Do Loop', so Anre is RIGHT, there is no infinite Do Loop involved.

You are just being sloppy.

Here is the context of the mistake that I was responding to:

On 4/2/2022 6:58 PM, André G. Isaak wrote:
 > it is obvious that if your 'SHD' had not aborted the
 > simulation, it *would* have ultimately reached a final state.

"it *would* have ultimately reached a final state."
refers to the simulated input.


But a PROPERLY simulated input WOULD reach the final state.


Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

No so much


But that isn't what H^ is.

You are the one that continues to insist that only a UTM can properly simulate an input and that this simulated input must never be aborted or it becomes an incorrect simulation.



Right, but that doesn't mean you put a UTM there. H^ has a copy of H there (that you are calling embedded_H), so that is wha must be there.


But (according to you) H cannot be there or it will get the simulation wrong.

You are confusing the REQUIREMENTS of H with the algorithm H itself.

H <M> w needs to go to Qy if M w (or UTM <M> w) will halt and Qn if it doesn't, but that doesn't mean you get to put a UTM there, because a UTM doesn't meet the requirements, since it doesn't give the non-halting answer in finite time.

H is NOT 'Computationally Equivalent' to a UTM, because its behavior IS DIFFERENT.


H simulates finite sequences of configurations 100% exactly the same way that a UTM does because it invokes its own internal UTM one step at a time.

For the non-halting case, UTM will be non-halting, but H MUST be a 'halting computation' that goes to the 'Non-Halting Answer State'

These are DIFFERENT.

Is that the confusion what you wasted 17 years on?

We need to put the actual algorithm of H at the point that was prevously labled Qx (Linz names it HQ0 to distinguish it from the Q0 starting

No he did not: https://www.liarparadox.org/Linz_Proof.pdf

I remembereded his notation wrong. He call the state that is the beginning of the orignial code for H starting at q0 with the name H^q0

Which is why the string execution paths are listed as:

q0 wM -> H^q0 wM wM -> H^ infinity [If M applied to wM Halts]
q0 wM -> H^q0 wM wM -> H^y1Qny2 [If M applied to wM never halts]


My latest notation is much better than his:

Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Ĥ.q0 ⟨Ĥ⟩ ⊢* H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qn

Except that you drop the conditions


The appended infinite loop is never reached, thus extraneous.

The behavior of the executed Ĥ applied to ⟨Ĥ⟩ still depends depends conditionally on the behavior of H. This same reasoning applies to UTM ⟨Ĥ⟩ ⟨Ĥ⟩.

The behavior of the input ⟨Ĥ⟩ ⟨Ĥ⟩ simulated by H cannot depend conditionally on anything because it remains stuck in infinitely nested simulation until aborted.


state of H^), so that is what is in the representation of H^ there, the representation of its copy of H, and that is what a UTM will see, and presumably what your SHD will see.

The only point we pull out the UTM, is when AS DESIGNERS/TESTERS we want to check if the answer that H gave was correct.

And, we are effectively doing that when we are just running H^ applied to <H^>, as that is BY DEFINITION the behavior of UTM <H^> <H^>.

Yes, one (perhaps you will call UNFAIR) aspect of this is that H can't just evaluate the requirements to get the right answer, but you should know from any actually useful programming that rarely can you just code 'The Requirements' to implement the program, as if you could, then there would be no need to hire a programmer.

H can get SOME non-halting cases with a list of patterns, but it turns out that there is no finite exhaustive list of patterns, which is why a Halt Decider is not possible (or the answer WOULD be to just use a SHD with that list of patterns).

All patterns are entirely comprised of a small set of elemental patterns this is very well known stuff.

Nope. Care to provide a reference that lists a COMPLETE set of non-halting patterns?

Yes, there are a number of basic loops that can become unending, but no COMPLETE list (as has been proven).


You are dumber than a box of rocks, this complete list is universally known as common knowledge.

An algorithm is made up of three basic building blocks:
sequencing, selection, and iteration.

An algorithm is made up of three basic building blocks:
sequencing, selection, and iteration.

An algorithm is made up of three basic building blocks:
sequencing, selection, and iteration.


An algorithm is made up of three basic building blocks: sequencing, selection, and iteration.

https://www.khanacademy.org/computing/ap-computer-science-principles/algorithms-101/building-algorithms/a/the-building-blocks-of-algorithms

So, what does that prove.

Simple blocks can create enormous complexity.

That also isn't really a complete definition, maybe right for a 101 level introductory course, but not for rigorous application.


If THAT is you source for Computer Science Definitions no wonder you are so messed up.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


Subject: Re: Refuting the Peter Linz Halting Problem Proof V6 [ honest dialogue is mandated ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Followup: comp.theory
Date: Sun, 3 Apr 2022 16:55 UTC
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On 4/3/2022 10:24 AM, olcott wrote:
On 4/3/2022 9:10 AM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/2/2022 6:57 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 4/2/2022 6:07 PM, Ben Bacarisse wrote:

Here's where we are.  After 17 years work on this you can not even say
what string must be passed to H so that H can tell us whether or not Ĥ
applied to ⟨Ĥ⟩ halts, and you won't say what state H.q0 ⟨Ĥ⟩ ⟨Ĥ⟩
transitions to even though everyone is happy to accept that
     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
from which the answer is easily obtained.  It would be extraordinary but
we know what's going on...

I did figure out the correct analysis of that.
The problem is that the result of the analysis is so counter-intuitive
that you will never pay enough attention to see that it is correct.

Good excuse not to say what string must be passed to H so that H can
tell us whether or not Ĥ applied to ⟨Ĥ⟩ halts and what state H.q0 ⟨Ĥ⟩
⟨Ĥ⟩ transitions to!

Keep not answering.  What else can you do?

If you want an honest dialogue you will go through the trouble of
working through mutual agreement on my key points...

Mutual agreement on nonsense can never form the basis of any reasonable
discussion.

That you can only dogmatically say that it is nonsense and not point out any error really seems to prove that you have no idea what I am saying because of your lack of technical skill.

I am not saying this as any denigrating put down. I am saying this so that we will be able to derive a common language and be able to have an effective dialogue.

You have been my longest reviewer and some aspects of your reviews have been very helpful. The key question that you ask is very important. Until we establish a "lingua franca" (common language) between us this dialogue cannot effectively occur.

KEY MISSING PIECE IN THE DIALOGUE
The key missing piece in all of these dialogues is 100% perfectly and exactly does it mean for a halt decider to compute the mapping from its input finite strings to its own final states on the basis of the actual   behavior actually specified by these finite strings.


--
Copyright 2022 Pete Olcott

"Talent hits a target no one else can hit;
  Genius hits a target no one else can see."
  Arthur Schopenhauer


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