On 12/19/2021 10:59 AM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 12/18/2021 5:06 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 12/17/2021 6:43 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 12/17/2021 3:25 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>>>> A function f with domain D is said to be Turing-computable
>>>>>>>> or just computable if there exists some Turing machine
>>>>>>>> M = (Q, Σ, Γ, δ, q0, □, F) such that q0 w ⊢* Mqff(w), qf ∈ F
>>>>>>>> for all w ∈ D (Linz:1990:243)
>>>>>>>
>>>>>>> We all know your cut and paste skills are top notch.
>>>>>>>
>>>>>>>> Olcott paraphrase of above machine definition: Machine M begins at
>>>>>>>> start state q0 on input w and transitions to qf as a function of input
>>>>>>>> w.
>>>>>>>
>>>>>>> But that's not a correct paraphrase. As for the jumble of words that is
>>>>>>> "the computable function copy of H at Ĥ.qx", well, I despair.
>>>>>>
>>>>>> I take this to mean that you have no idea what the behavior of Ĥ
>>>>>> applied to ⟨Ĥ⟩ would be:
>>>>
>>>>> You should, in general, take me to mean what I say, with as little
>>>>> interpretation as you can muster. Your track record in paraphrasing me
>>>>> is abysmal. Is my claim that I despair at your jumble of words in some
>>>>> way unclear?
>>>>
>>>> Since this is equivalent to Linz notation you can't say that my
>>>> question is not clear?
>>>
>>> You are loosing the plot again. I called out your stupid paraphrase of my
>>> words by asking what was unclear about what I'd said.
>>> You, however, are always unclear:
>>>
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy
>>>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>
>>>> What would the behavior of Ĥ applied to ⟨Ĥ⟩ be if we assume that Ĥ.qx
>>>> acts exactly as if it was UTM(⟨Ĥ⟩, ⟨Ĥ⟩) ???
>>>
>>> Here you abuse Linz's hat notation by using it to mean something it does
>>> not. Now we have no idea what you mean by Ĥ in any posts form here on
>>> down. So much for you childish repetition, not so very long ago, that
>>> you always mean Linz's Ĥ when you write Ĥ. Not so anymore.
>>
>> I change one element of Ĥ and ask you to tell me what difference that
>> makes and you dodge because you simply don't know.
>
> If that makes you feel better, you go right ahead and think that. But
> remember, from now on you have to say which Ĥ you mean when you write
> Ĥ. And you will have to say what H is too since the above Ĥ is built
> from a H quite unlike Linz's H.
>
> I'll leave you with the proof in which you have still been unable to
> find any flaw. (Yes, I know you point at the last line and SHOUT, but
> that's not how mathematics works.)
>
> If a TM, H, existed such that
>
> Hq0 <M> s ⊦* Hqy if M applied to s halts, and
> Hq0 <M> s ⊦* Hqn if M applied to s does not halt,
>
> we could construct from it an H' such that
>
> H'q0 <M> s ⊦* oo if M applied to s halts, and
> H'q0 <M> s ⊦* H'qn if M applied to s does not halt.
>
> And from that we could construct an Ĥ such that
>
> Ĥq0 <M> ⊦* Ĥqx <M> <M> ⊦* oo if M applied to <M> halts, and
> Ĥq0 <M> ⊦* Ĥqx <M> <M> ⊦* Ĥqn if M applied to <M> does not halt.
>
> Setting M to Ĥ, so <M> becomes <Ĥ>, we see that the existence of H (as
> Linz defines it) logically entails that a TM Ĥ that does this
>
> Ĥq0 <Ĥ> ⊦* Ĥqx <Ĥ> <Ĥ> ⊦* oo if Ĥ applied to <Ĥ> halts, and
> Ĥq0 <Ĥ> ⊦* Ĥqx <Ĥ> <Ĥ> ⊦* Ĥqn if Ĥ applied to <Ĥ> does not halt.
>

What is the sequence of configurations specified by Ĥ applied to ⟨Ĥ⟩
when we assume that the Linz H is a simulating halt decider and that Ĥ
does not have the infinite loop appended to its Ĥ.qy state?

Because H is a simulating halt decider this hypothetical definition of Ĥ
defines H as a pure simulator, thus proving that simulating halt decider
H must abort the simulation of its input.

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

--
Copyright 2021 Pete Olcott

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

On 12/19/2021 12:13 PM, Richard Damon wrote:
> On 12/19/21 12:40 PM, olcott wrote:
>> On 12/19/2021 10:59 AM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 12/18/2021 5:06 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 12/17/2021 6:43 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 12/17/2021 3:25 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>>>> A function f with domain D is said to be Turing-computable
>>>>>>>>>> or just computable if there exists some Turing machine
>>>>>>>>>> M = (Q, Σ, Γ, δ, q0, □, F) such that q0 w ⊢* Mqff(w), qf ∈ F
>>>>>>>>>> for all w ∈ D (Linz:1990:243)
>>>>>>>>>
>>>>>>>>> We all know your cut and paste skills are top notch.
>>>>>>>>>
>>>>>>>>>> Olcott paraphrase of above machine definition: Machine M
>>>>>>>>>> begins at
>>>>>>>>>> start state q0 on input w and transitions to qf as a function
>>>>>>>>>> of input
>>>>>>>>>> w.
>>>>>>>>>
>>>>>>>>> But that's not a correct paraphrase.  As for the jumble of
>>>>>>>>> words that is
>>>>>>>>> "the computable function copy of H at Ĥ.qx", well, I despair.
>>>>>>>>
>>>>>>>> I take this to mean that you have no idea what the behavior of Ĥ
>>>>>>>> applied to ⟨Ĥ⟩ would be:
>>>>>>
>>>>>>> You should, in general, take me to mean what I say, with as little
>>>>>>> interpretation as you can muster.  Your track record in
>>>>>>> paraphrasing me
>>>>>>> is abysmal.  Is my claim that I despair at your jumble of words
>>>>>>> in some
>>>>>>> way unclear?
>>>>>>
>>>>>> Since this is equivalent to Linz notation you can't say that my
>>>>>> question is not clear?
>>>>>
>>>>> You are loosing the plot again.  I called out your stupid
>>>>> paraphrase of my
>>>>> words by asking what was unclear about what I'd said.
>>>>> You, however, are always unclear:
>>>>>
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy
>>>>>>      Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>>>>>
>>>>>> What would the behavior of Ĥ applied to ⟨Ĥ⟩ be if we assume that Ĥ.qx
>>>>>> acts exactly as if it was UTM(⟨Ĥ⟩, ⟨Ĥ⟩) ???
>>>>>
>>>>> Here you abuse Linz's hat notation by using it to mean something it
>>>>> does
>>>>> not.  Now we have no idea what you mean by Ĥ in any posts form here on
>>>>> down.  So much for you childish repetition, not so very long ago, that
>>>>> you always mean Linz's Ĥ when you write Ĥ.  Not so anymore.
>>>>
>>>> I change one element of Ĥ and ask you to tell me what difference that
>>>> makes and you dodge because you simply don't know.
>>>
>>> If that makes you feel better, you go right ahead and think that.  But
>>> remember, from now on you have to say which Ĥ you mean when you write
>>> Ĥ.  And you will have to say what H is too since the above Ĥ is built
>>> from a H quite unlike Linz's H.
>>>
>>> I'll leave you with the proof in which you have still been unable to
>>> find any flaw.  (Yes, I know you point at the last line and SHOUT, but
>>> that's not how mathematics works.)
>>>
>>> If a TM, H, existed such that
>>>
>>>    Hq0 <M> s ⊦* Hqy    if M applied to s halts, and
>>>    Hq0 <M> s ⊦* Hqn    if M applied to s does not halt,
>>>
>>> we could construct from it an H' such that
>>>
>>>    H'q0 <M> s ⊦* oo      if M applied to s halts, and
>>>    H'q0 <M> s ⊦* H'qn    if M applied to s does not halt.
>>>
>>> And from that we could construct an Ĥ such that
>>>
>>>    Ĥq0 <M> ⊦* Ĥqx <M> <M> ⊦* oo     if M applied to <M> halts, and
>>>    Ĥq0 <M> ⊦* Ĥqx <M> <M> ⊦* Ĥqn    if M applied to <M> does not halt.
>>>
>>> Setting M to Ĥ, so <M> becomes <Ĥ>, we see that the existence of H (as
>>> Linz defines it) logically entails that a TM Ĥ that does this
>>>
>>>    Ĥq0 <Ĥ> ⊦* Ĥqx <Ĥ> <Ĥ> ⊦* oo     if Ĥ applied to <Ĥ> halts, and
>>>    Ĥq0 <Ĥ> ⊦* Ĥqx <Ĥ> <Ĥ> ⊦* Ĥqn    if Ĥ applied to <Ĥ> does not halt.
>>>
>>
>> What is the sequence of configurations specified by Ĥ applied to ⟨Ĥ⟩
>> when we assume that the Linz H is a simulating halt decider and that Ĥ
>> does not have the infinite loop appended to its Ĥ.qy state?
>>
>
> And what do you mean by H is a simulating Halt Decider?
>
> Have you presumed (as an unproven fact) that H is correct?
>

I can (and will shortly) conclusively prove that pure function H
correctly detects that P specifies infinitely nested simulation.

// Simplified Linz(1990) Ĥ
// and Strachey(1965) P
void P(ptr x)
{ if (H(x, x))
HERE: goto HERE;
}

I can (and will shortly) conclusively prove that pure function H
correctly detects that P specifies infinitely nested simulation.
This same proof also now works with the following:

// Linz(1990) Ĥ
void P(ptr x)
{ ptr y = Copy(x);
if (H(x, y))
HERE: goto HERE;
}

> It really sounds like you have lost track of the problem you were trying
> to solve, and are no longer trying to prove the existence of H by giving
> an example, but just ASSUMING it exists and then talking about it, and
> IGNORING all the contractions its creates or blaming them on H^.
>
>> Because H is a simulating halt decider this hypothetical definition of
>> Ĥ defines H as a pure simulator, thus proving that simulating halt
>> decider H must abort the simulation of its input.
>
> But a 'Pure Simulator' can't be a Halt Decider, as it can't decide a
> non-halting pattern, BY DEFINITION.
>
> BY DEFINITION, A 'Pure Simulator' will run forever when its input
> represents a non-halting computation, but a Halt Decider, BY DEFINITION,
> must return an answer in FINITE time, thus NO 'Pure Simulator' can be a
> correct Halt Decider.
>
>
>>
>>       Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy
>>       Ĥ.q0 ⟨Ĥ⟩ ⊢* UTM ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
>>
>>
>>
>
> FAIL. You just proved your Halt Decider never answer for this Problem.

--
Copyright 2021 Pete Olcott

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

On 12/19/2021 4:25 PM, Malcolm McLean wrote:
> On Sunday, 19 December 2021 at 19:04:11 UTC, Jeff Barnett wrote:
>> On 12/19/2021 11:44 AM, olcott wrote:
>>> I can (and will shortly) conclusively prove that pure function H
>>> correctly detects that P specifies infinitely nested simulation.
>> Simple question: Since virtually everyone who has studied this problem
>> knows, simulation is not a feasible way to build a halt decider
>> (assuming one could be built). The proof you are attacking as well as
>> others I am aware of are not based on simulation. Why do insist that
>> simulation is the way to go?
>>
> I think I can answer this for PO.
>
> The first observation is that a simulating halt decider goes into an
> infinite series of nested simulations if fed H_Hat. So this leads to the
> question, can we somehow exploit this to get round the Linz trap?
>
> The answer is "no". But in answering that question, we set up a system
> in which the simulating halt decider must get the wrong answer, for reasons
> which have nothing to do with LInz's "invert the result" step. So have
> we broken new ground? Can we maybe say that the result is correct,
> because the simulation "aborts" instead of "halting", and if it didn't
> "abort" it would indeed run forever, so it must have correctly "aborted".

Thanks for providing a reasonable unbiased critique.

Because computable functions are only accountable for their actual
inputs when H computes the halt status of its input and finds that the
pure simulation of this input never stops running it is correct to abort
this simulation and report non-halting.

As with all inputs to all simulating halt deciders, the correct halt
status basis is whether or not the pure simulation of its input ever
stops running without being aborted.

--
Copyright 2021 Pete Olcott

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

On 12/19/2021 4:25 PM, Malcolm McLean wrote:
> On Sunday, 19 December 2021 at 19:04:11 UTC, Jeff Barnett wrote:
>> On 12/19/2021 11:44 AM, olcott wrote:
>>> I can (and will shortly) conclusively prove that pure function H
>>> correctly detects that P specifies infinitely nested simulation.
>> Simple question: Since virtually everyone who has studied this problem
>> knows, simulation is not a feasible way to build a halt decider
>> (assuming one could be built). The proof you are attacking as well as
>> others I am aware of are not based on simulation. Why do insist that
>> simulation is the way to go?
>>
> I think I can answer this for PO.
>
> The first observation is that a simulating halt decider goes into an
> infinite series of nested simulations if fed H_Hat. So this leads to the
> question, can we somehow exploit this to get round the Linz trap?
>

The HP counter-examples have always assumed that both answers that the
corresponding halt decider could provide, halts and does not halt would
be the wrong answer.

THIS BY ITSELF IS A BRAND NEW IDEA
These same counter-examples provided to a simulating halt decider
specify the non-halting behavior of infinitely nested simulation.

> The answer is "no". But in answering that question, we set up a system
> in which the simulating halt decider must get the wrong answer, for reasons

Everyone that ever said that it must get the wrong answer did not
understand the computer science well enough.

A halt decider only computes the mapping of its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to an
accept / reject state. It must do this entirely on the basis of the
actual behavior specified by these actual inputs.

As long as no amount of simulation by the simulating halt decider would
ever cause the simulated input to reach its final state we know by the
Linz definition of halting that this input specifies a non halting
computation.

Computation that halts: the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)

As long as the input to a halt decider specifies a non halting
computation then

this by itself is an entirely sufficient condition
this by itself is an entirely sufficient condition
this by itself is an entirely sufficient condition
this by itself is an entirely sufficient condition

for the halt decider to transition to its reject state.

> which have nothing to do with LInz's "invert the result" step. So have
> we broken new ground? Can we maybe say that the result is correct,
> because the simulation "aborts" instead of "halting", and if it didn't
> "abort" it would indeed run forever, so it must have correctly "aborted".

The verified fact that the input specifies a non-halting computation to
every simulating halt decider provides these halt deciders the entirely
sufficient basis for to transition to their reject state.

--
Copyright 2021 Pete Olcott

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

On 1/5/2022 8:46 AM, Malcolm McLean wrote:
> On Wednesday, 5 January 2022 at 14:31:17 UTC, olcott wrote:
>> On 1/5/2022 5:01 AM, Julio Di Egidio wrote:
>>> On Wednesday, 5 January 2022 at 06:34:32 UTC+1, richar...@gmail.com wrote:
>>>> On 1/4/22 11:15 PM, olcott wrote:
>>>>> On 1/4/2022 10:09 PM, Richard Damon wrote:
>>>>>> On 1/4/22 10:43 PM, olcott wrote:
>>>>>>> On 1/4/2022 8:04 PM, Richard Damon wrote:
>>>>>>>> On 1/4/22 7:29 PM, olcott wrote:
>>>>>>>>> On 12/19/2021 4:25 PM, Malcolm McLean wrote:
>>>>>>>>>> On Sunday, 19 December 2021 at 19:04:11 UTC, Jeff Barnett wrote:
>>>>>>>>>>> On 12/19/2021 11:44 AM, olcott wrote:
>>>
>>>> FAIL.
>>>
>>> STOP FEEDING THE TROLLS: you fucking retarded spamming pieces of shit are the true disgrace here.
>>>
>>> *Plonk*
>>>
>>> Julio
>> Dumb Bunny !!!
>>
>> Plonk is a Usenet jargon term for adding a particular poster to one's
>> kill file so that poster's future postings are completely ignored.
>> https://news.ycombinator.com/item?id=26890646
>>
> Julio has added Richard to his kill file, because he is annoyed by his responses
> to you. The "plonk" suggests that other people do the same.

No not at all, Julio has very persistently (over the years) proved that
he has no idea what the term "Plonk" means. He continues to respond to
people that he has "Plonked" after he has "Plonked" them.

> I will not be doing so. Though I do agree that Richard doesn't have anything much
> new to say that hasn't been said many times already, I'll leave it to his judgement
> whether it is worth repeating essentially the same point many times over.
>

This aspect of your assessment is incorrect:
This aspect of your assessment is incorrect:
This aspect of your assessment is incorrect:

On 12/19/2021 4:25 PM, Malcolm McLean wrote:
> But in answering that question, we set up a system in which
> the simulating halt decider must get the wrong answer,

Everyone that ever said that it must get the wrong answer did not
understand the computer science well enough.

A halt decider only computes the mapping of its inputs ⟨Ĥ⟩ ⟨Ĥ⟩ to an
accept / reject state. It must do this entirely on the basis of the
actual behavior specified by these actual inputs.

As long as no amount of simulation by the simulating halt decider would
ever cause the simulated input to reach its final state we know by the
Linz definition of halting that this input specifies a non halting
computation.

Computation that halts: the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)

As long as the input to a halt decider specifies a non halting
computation then

this by itself is an entirely sufficient condition
this by itself is an entirely sufficient condition
this by itself is an entirely sufficient condition
this by itself is an entirely sufficient condition

for the halt decider to transition to its reject state.

--
Copyright 2021 Pete Olcott

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

On 12/19/2021 4:25 PM, Malcolm McLean wrote:
> On Sunday, 19 December 2021 at 19:04:11 UTC, Jeff Barnett wrote:
>> On 12/19/2021 11:44 AM, olcott wrote:
>>> I can (and will shortly) conclusively prove that pure function H
>>> correctly detects that P specifies infinitely nested simulation.
>> Simple question: Since virtually everyone who has studied this problem
>> knows, simulation is not a feasible way to build a halt decider
>> (assuming one could be built). The proof you are attacking as well as
>> others I am aware of are not based on simulation. Why do insist that
>> simulation is the way to go?
>>
> I think I can answer this for PO.
>
> The first observation is that a simulating halt decider goes into an
> infinite series of nested simulations if fed H_Hat. So this leads to the
> question, can we somehow exploit this to get round the Linz trap?
>

You are the only reviewer that has ever made any attempt at an honest
dialogue. (Except possibly Kaz) That one party of a conversation has no
intention of any honest dialogue is proven by the fact that there are
persistently zero elements of mutual agreement by one of the parties.

Every time that I effectively prove my point those respondents that have
no interest what-so-ever in any honest dialogue always change the
subject to another basis of rebuttal.

When Ben responded to your point he studiously totally bypassed the
point by nitpicking at the use of terminology.

Sapir–Whorf hypothesis
language determines thought and that linguistic categories limit and
determine cognitive categories.
https://en.wikipedia.org/wiki/Linguistic_relativity

In order to talk about TMs that base their halt status decision on the
behavior of simulating N steps of their input until this input either
halts on its own or the TM recognizes an infinitely behavior pattern we
must have a term.

Since the term "decider" must halt on all inputs, the term decider may
not be perfectly apt unless we qualify it.

The most straight forward way to qualify such a term would be that a TM
is a simulating halt decider that computes the halt status of a limited
domain of finite string pairs.

H computes the function of mapping input pairs to an accept / reject
state on the basis of simulating N steps of this input pair until this
simulated input reaches its final state or H has recognize an infinite
behavior pattern. Because the purpose of H is limited to showing how the
conventional HP counter-examples would be decided the function that H
computes its limited to one element of these counter-example inputs.

> The answer is "no". But in answering that question, we set up a system
> in which the simulating halt decider must get the wrong answer, for reasons
> which have nothing to do with LInz's "invert the result" step. So have
> we broken new ground? Can we maybe say that the result is correct,
> because the simulation "aborts" instead of "halting", and if it didn't
> "abort" it would indeed run forever, so it must have correctly "aborted".

computation that halts… the Turing machine will halt whenever it enters
a final state. (Linz:1990:234)

As soon as H detects that the simulation of P cannot possibly ever reach
the final state of P then P meets the Linz definition of a computation
that does not halt.

Once H knows that P specifies a non-halting computation this is entirely
sufficient for H to transition to its final reject state.

Not_Halting(P) → H.Reject(P)

https://en.wikipedia.org/wiki/Necessity_and_sufficiency
As long as the input to H specifies a computation that never halts H is
always correct to transition to its reject state.

--
Copyright 2021 Pete Olcott

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

On 1/5/2022 4:15 PM, Richard Damon wrote:
> On 1/5/22 4:01 PM, olcott wrote:
>> On 1/5/2022 2:55 PM, Richard Damon wrote:
>>> On 1/5/22 3:41 PM, olcott wrote:
>>>> On 1/5/2022 2:36 PM, Richard Damon wrote:
>>>>> On 1/5/22 3:03 PM, olcott wrote:
>>>>>> On 1/5/2022 1:54 PM, Richard Damon wrote:
>>>>>>> On 1/5/22 1:00 PM, olcott wrote:
>>>>>>>> On 1/5/2022 10:29 AM, Richard Damon wrote:
>>>>>>>>> On 1/5/22 11:06 AM, olcott wrote:
>>>>>>>>>> On 12/19/2021 4:25 PM, Malcolm McLean wrote:
>>>>>>>>>>> On Sunday, 19 December 2021 at 19:04:11 UTC, Jeff Barnett wrote:
>>>>>>>>>>>> On 12/19/2021 11:44 AM, olcott wrote:
>>>>>>>>>>>>> I can (and will shortly) conclusively prove that pure
>>>>>>>>>>>>> function H
>>>>>>>>>>>>> correctly detects that P specifies infinitely nested
>>>>>>>>>>>>> simulation.
>>>>>>>>>>>> Simple question: Since virtually everyone who has studied
>>>>>>>>>>>> this problem
>>>>>>>>>>>> knows, simulation is not a feasible way to build a halt decider
>>>>>>>>>>>> (assuming one could be built). The proof you are attacking
>>>>>>>>>>>> as well as
>>>>>>>>>>>> others I am aware of are not based on simulation. Why do
>>>>>>>>>>>> insist that
>>>>>>>>>>>> simulation is the way to go?
>>>>>>>>>>>>
>>>>>>>>>>> I think I can answer this for PO.
>>>>>>>>>>>
>>>>>>>>>>> The first observation is that a simulating halt decider goes
>>>>>>>>>>> into an
>>>>>>>>>>> infinite series of nested simulations if fed H_Hat. So this
>>>>>>>>>>> leads to the
>>>>>>>>>>> question, can we somehow exploit this to get round the Linz
>>>>>>>>>>> trap?
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> You are the only reviewer that has ever made any attempt at an
>>>>>>>>>> honest dialogue. (Except possibly Kaz) That one party of a
>>>>>>>>>> conversation has no intention of any honest dialogue is proven
>>>>>>>>>> by the fact that there are persistently zero elements of
>>>>>>>>>> mutual agreement by one of the parties.
>>>>>>>>>>
>>>>>>>>>> Every time that I effectively prove my point those respondents
>>>>>>>>>> that have no interest what-so-ever in any honest dialogue
>>>>>>>>>> always change the subject to another basis of rebuttal.
>>>>>>>>>>
>>>>>>>>>> When Ben responded to your point he studiously totally
>>>>>>>>>> bypassed the point by nitpicking at the use of terminology.
>>>>>>>>>>
>>>>>>>>>> Sapir–Whorf hypothesis
>>>>>>>>>> language determines thought and that linguistic categories
>>>>>>>>>> limit and determine cognitive categories.
>>>>>>>>>> https://en.wikipedia.org/wiki/Linguistic_relativity
>>>>>>>>>>
>>>>>>>>>> In order to talk about TMs that base their halt status
>>>>>>>>>> decision on the behavior of simulating N steps of their input
>>>>>>>>>> until this input either halts on its own or the TM recognizes
>>>>>>>>>> an infinitely behavior pattern we must have a term.
>>>>>>>>>>
>>>>>>>>>> Since the term "decider" must halt on all inputs, the term
>>>>>>>>>> decider may not be perfectly apt unless we qualify it.
>>>>>>>>>
>>>>>>>>> But since the Theorem that you want to actually talk about is
>>>>>>>>> based on the term Decider, you can't change the meaning of the
>>>>>>>>> term or use a slightly different term and still be talking
>>>>>>>>> about THAT Theorem.
>>>>>>>>>
>>>>>>>>> You don't get to 'change' the meaning of the terms in the
>>>>>>>>> Theorem, no matter how inconvenient that makes things for you.
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> The most straight forward way to qualify such a term would be
>>>>>>>>>> that a TM is a simulating halt decider that computes the halt
>>>>>>>>>> status of a limited domain of finite string pairs.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> And to do the job you claim to be doing that domain MUST
>>>>>>>>> include the string that represents the description of the H^
>>>>>>>>> machine build on your exact H being applied to that string.
>>>>>>>>>
>>>>>>>>
>>>>>>>> Yes, that is the only element of the domain of the simulating
>>>>>>>> halt decider hat I am referring to.
>>>>>>>>
>>>>>>>>>> H computes the function of mapping input pairs to an accept /
>>>>>>>>>> reject state on the basis of simulating N steps of this input
>>>>>>>>>> pair until this simulated input reaches its final state or H
>>>>>>>>>> has recognize an infinite behavior pattern. Because the
>>>>>>>>>> purpose of H is limited to showing how the conventional HP
>>>>>>>>>> counter-examples would be decided the function that H computes
>>>>>>>>>> its limited to one element of these counter-example inputs.
>>>>>>>>>
>>>>>>>>> And to meet the REQUIREMENTS of the Theory, this accept/reject
>>>>>>>>> needs to match the ACTUAL BEHAVIOR of that H^ machine built
>>>>>>>>> from the H you are claiming to be correct applied to its
>>>>>>>>> description.
>>>>>>>>>
>>>>>>>>
>>>>>>>> If it is the case that the pure simulation of ⟨Ĥ⟩ ⟨Ĥ⟩ by
>>>>>>>> embedded_H would never reach the final state of this simulated
>>>>>>>> input then it would be  known with 100% perfectly justified
>>>>>>>> complete logical certainty that this input specifies a
>>>>>>>> non-halting computation.
>>>>>>>
>>>>>>> If it is the case that simulation of <H^> <H^> by H WILL never
>>>>>>> reach the final state, then this MEANS that since H^ USES H, that
>>>>>>> H, which is IDENTICAL to this one, must never abort its
>>>>>>> simulation and go to H.Qn, thus you premise is that H never goes
>>>>>>> to H.Qn.
>>>>>>>
>>>>>>
>>>>>> This might seem this way to a brain dead moron that can't begin to
>>>>>> imagine that a simulating halt decider could recognize that
>>>>>> something as simple as an infinite loop would never stop running
>>>>>> unless this simulating halt decider actually waited an infinite
>>>>>> amount of time to verify that the infinite loop never stops running.
>>>>>>
>>>>>> Are you really a brain dead moron or merely pretending to be one
>>>>>> as a vicious head game?
>>>>>
>>>>> You may see me as 'Brain Dead', but you are the one who seems to
>>>>> not be able to read what people say and make an appropriate response.
>>>>>
>>>>> If you think this task is so easy, just show your code that does it.
>>>>>
>>>>
>>>> One need not see any code when one knows that when the invocation of
>>>> a function results in the invocation of this same function with the
>>>> same inputs that this proves infinite behavior.
>>>>
>>>>  From what I recall this was your own idea that you now deceitfully
>>>> deny.
>>>
>>> Yes, INVOCATION of the same function in a CALL chain will generate
>>> infinite behavior,
>>>
>>> If H 'calls' its input (maybe by UTM Simulation) this would be the
>>> behavior, but such an H then has lost the ability to 'abort' the
>>> 'simulation' to return the answer.
>>>
>>
>> You and Ben both deny that a halt decider could have simulating
>> ability, this is so ridiculously stupid that it must be flat out
>> dishonesty.
>
> LIE!!!
>
> I never said that, please show where I did or retract it.
>
> Maybe I should have a solicitor pay you a call with a defamation suit.
>
> Maybe if you had to try to explain your logic to a jury under penalty of
> perjury you would shape up.
>
>>
>> Since I have a fully operational halt decider that does have
>> simulating ability this proves that you are a damned liar.
>
> But it doesn't give the right answer, so YOU are the damned liar.
>

On 1/6/2022 5:58 AM, Malcolm McLean wrote:
> On Wednesday, 5 January 2022 at 23:22:33 UTC, olcott wrote:
>> On 1/5/2022 5:10 PM, Richard Damon wrote:
>>>
>>> No, YOU do that, you have several different things that you call H, with
>>> different behaviors.
>>>
>> It is not that hard.
>> (a) H directly executes its input
>> (b) H simulates its input
>> (c) H watches the simulation of its input looking for infinite behavior
>> patterns.
>>
> It does seem when first considering the problem that this approach
> should work. If the halt decider fails on some input, then that just
> indicates that the infinite behaviour pattern detector needs to be more
> sophisticated.
>
> But when you really get into it, you see that in fact that's a false
> impression. Even some fairly simple Turing machines will defeat any
> infinite behaviour pattern detector you can devise.
>

Try and provide an example.
My goal was to defeat the conventional HP counter-examples.
It seems that I have done that.

--
Copyright 2021 Pete Olcott

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