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* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ keyolcott
+- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
+- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
 `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |+* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Malcolm McLean
    ||`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    || `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Malcolm McLean
    ||  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    ||   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    ||   |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    ||   | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    ||   |  `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    ||   `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    | |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    | |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    | |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    | |     `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |     +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |     `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |      +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |      |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |      | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |      |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |      |   `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |      `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |       `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |   |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |+- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
    |   |        |    |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |     `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Jeff Barnett
    |   |        |    |      |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Jeff Barnett
    |   |        |    |      |   |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   | `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |  `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |     `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |      `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |       +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |      |   |       |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |      |   |       `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |      |   `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |      `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |       `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |        `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |         `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |          +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |          `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |           `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |            |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Dennis Bush
    |   |        |    |            | |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
    |   |        |    |            | | |`* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | | | +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Python
    |   |        |    |            | | | `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |            | | `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |            | +* Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse
    |   |        |    |            | |`- Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |            | `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Richard Damon
    |   |        |    |            `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |             +- Refuting the Peter Linz Halting Problem Proof --- Version(10) [Malcolm McLean
    |   |        |    |             `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |              `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |               `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    |                `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   |        |    |                 `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        |    `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [olcott
    |   |        `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [André G. Isaak
    |   `* Refuting the Peter Linz Halting Problem Proof --- Version(10) [Andy Walker
    `- Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piecBen Bacarisse

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Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<v6idnaCJifSVTtT_nZ2dnUU7_8zNnZ2d@giganews.com>

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 by: olcott - Sun, 3 Apr 2022 17:01 UTC

KEY MISSING PIECE IN THE DIALOGUE (that all reviewers get incorrectly)

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 this finite string pair.

Halting problem undecidability and infinitely nested simulation (V4)
https://www.researchgate.net/publication/359349179_Halting_problem_undecidability_and_infinitely_nested_simulation_V4

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<t2ck4o$1jtf$1@gioia.aioe.org>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
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 by: Python - Sun, 3 Apr 2022 17:05 UTC

Peter Olcott wrote:
> KEY MISSING PIECE IN THE DIALOGUE (that all reviewers get incorrectly)
>
> The key missing piece in all of these dialogues is ...

There are two of them.

Your mental sanity and your intellectual integrity.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<4fl2K.601639$oF2.47701@fx10.iad>

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 by: Richard Damon - Sun, 3 Apr 2022 17:58 UTC

On 4/3/22 1:01 PM, olcott wrote:
> KEY MISSING PIECE IN THE DIALOGUE (that all reviewers get incorrectly)
>
> 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 this finite string pair.
>
> Halting problem undecidability and infinitely nested simulation (V4)
> https://www.researchgate.net/publication/359349179_Halting_problem_undecidability_and_infinitely_nested_simulation_V4
>
>

The KEY to that is that the mapping is perfectly defined by the
definitions of the field.

H applied to <M> w needs to go to Qy iff M applied to w will reach a
final state in a finite number of steps, and to go to Qn iff M applied
to w will never reach a final state even in an unbounded number of steps.

This is the DEFINITION of the Mapping that a Halt Decider MUST be trying
to compute.

Yes, this doesn't specify HOW the decider does its job, that is not the
role of the problem, How is a matter of the person attempting to solve
the problem. All that is specified is that the Decider needs to be a
Turing Machine.

One key point is that there is no promise that this mapping IS
computable, and in fact, the ultimate question of the problem is that
question, is Halting (or more precisely Non-Halting) a Computable Function.

Any argument that starts from the requirement that the definition needs
to define HOW to compute the results is inherently flawed, because that
it NOT the original or ultimate question. In fact, a significant part of
the question is whether it is possible at all.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<87sfquc7ns.fsf@bsb.me.uk>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
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 by: Ben Bacarisse - Sun, 3 Apr 2022 19:02 UTC

olcott <NoOne@NoWhere.com> writes:

> 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 this finite string pair.

You certainly have trouble understanding this so I will repeat my offer
to take you through a series of student exercises that I am confident
(provided you approach them with an open mind) will lead you to fully
understanding what this means.

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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 by: olcott - Sun, 3 Apr 2022 20:26 UTC

On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> 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 this finite string pair.
>
> You certainly have trouble understanding this so I will repeat my offer
> to take you through a series of student exercises that I am confident
> (provided you approach them with an open mind) will lead you to fully
> understanding what this means.
>

OK let's give it a shot. I know that I must be correct yet when you try
to explain these things using your words then this will probably result
in much greater mutual understanding, thus worth the effort. After we do
this I can explain my ideas using your own words.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<871qydde5e.fsf@bsb.me.uk>

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Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
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 by: Ben Bacarisse - Sun, 3 Apr 2022 21:56 UTC

olcott <NoOne@NoWhere.com> writes:

> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> 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 this finite string pair.
>> You certainly have trouble understanding this so I will repeat my offer
>> to take you through a series of student exercises that I am confident
>> (provided you approach them with an open mind) will lead you to fully
>> understanding what this means.
>
> OK let's give it a shot.

Great! Let's start with something that will force a lot of hidden
assumptions to be made explicit.

Q: Is the set of even numbers decidable? If so, specify the TM (call it
E) using Linz's notation (extended if you like). If not, why not?

(This may be slow to get started, because there is a lot low-level
things to iron out first.)

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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 by: olcott - Sun, 3 Apr 2022 22:07 UTC

On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> 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 this finite string pair.
>>> You certainly have trouble understanding this so I will repeat my offer
>>> to take you through a series of student exercises that I am confident
>>> (provided you approach them with an open mind) will lead you to fully
>>> understanding what this means.
>>
>> OK let's give it a shot.
>
> Great! Let's start with something that will force a lot of hidden
> assumptions to be made explicit.
>
> Q: Is the set of even numbers decidable? If so, specify the TM (call it
> E) using Linz's notation (extended if you like). If not, why not?
>
> (This may be slow to get started, because there is a lot low-level
> things to iron out first.)
>

That depends on how you define your terms. Any element of the set of
integers can be determined to be divisible by 2 with or without a
remainder.

Can every element of this set be enumerated in finite time?
No. Can the set of all even numbers be defined in finite space?
Yes through algorithmic compression.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<w-edndQxIuTUhtf_nZ2dnUU7_81g4p2d@giganews.com>

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 by: olcott - Sun, 3 Apr 2022 22:10 UTC

On 4/3/2022 5:07 PM, olcott wrote:
> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> 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 this finite string pair.
>>>> You certainly have trouble understanding this so I will repeat my offer
>>>> to take you through a series of student exercises that I am confident
>>>> (provided you approach them with an open mind) will lead you to fully
>>>> understanding what this means.
>>>
>>> OK let's give it a shot.
>>
>> Great!  Let's start with something that will force a lot of hidden
>> assumptions to be made explicit.
>>
>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>     E) using Linz's notation (extended if you like).  If not, why not?
>>
>> (This may be slow to get started, because there is a lot low-level
>> things to iron out first.)
>>
>
> That depends on how you define your terms. Any element of the set of
> integers can be determined to be divisible by 2 with or without a
> remainder.
>
> Can every element of this set be enumerated in finite time?
> No. Can the set of all even numbers be defined in finite space?
> Yes through algorithmic compression.

I forgot decidability is merely a membership algorithm, so yes.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
From: malcolm....@gmail.com (Malcolm McLean)
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 by: Malcolm McLean - Sun, 3 Apr 2022 22:25 UTC

On Sunday, 3 April 2022 at 23:10:26 UTC+1, olcott wrote:
> On 4/3/2022 5:07 PM, olcott wrote:
> > On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
> >> olcott <No...@NoWhere.com> writes:
> >>
> >>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
> >>>> olcott <No...@NoWhere.com> writes:
> >>>>
> >>>>> 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 this finite string pair.
> >>>> You certainly have trouble understanding this so I will repeat my offer
> >>>> to take you through a series of student exercises that I am confident
> >>>> (provided you approach them with an open mind) will lead you to fully
> >>>> understanding what this means.
> >>>
> >>> OK let's give it a shot.
> >>
> >> Great! Let's start with something that will force a lot of hidden
> >> assumptions to be made explicit.
> >>
> >> Q: Is the set of even numbers decidable? If so, specify the TM (call it
> >> E) using Linz's notation (extended if you like). If not, why not?
> >>
> >> (This may be slow to get started, because there is a lot low-level
> >> things to iron out first.)
> >>
> >
> > That depends on how you define your terms. Any element of the set of
> > integers can be determined to be divisible by 2 with or without a
> > remainder.
> >
> > Can every element of this set be enumerated in finite time?
> > No. Can the set of all even numbers be defined in finite space?
> > Yes through algorithmic compression.
> I forgot decidability is merely a membership algorithm, so yes.
>
Yes. The question is how the input is to be encoded. Whilst you could use
any encoding, the obvious options are a series of N "1"s followed by a blank,
or a number in binary, followed by a blank.
Next thing to do is to write the Turing machine that accepts an even number,
and rejects everything that isn't an even number. For simplicity, say that all
inputs will be either odd numbers or even numbers, and any input that isn't a
number can be disregarded.
Note that the numbers can be of arbitrary length, though not infinite length.
That's one of the nice things about Turing machines. Unlike C programs,
the machines tend to handle huge inputs without any modification.

You'll need some environment for specifying and test-running Turing machines
to get anywhere very fast. There must be a good online resource.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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 by: olcott - Sun, 3 Apr 2022 22:44 UTC

On 4/3/2022 5:25 PM, Malcolm McLean wrote:
> On Sunday, 3 April 2022 at 23:10:26 UTC+1, olcott wrote:
>> On 4/3/2022 5:07 PM, olcott wrote:
>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>> olcott <No...@NoWhere.com> writes:
>>>>
>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>
>>>>>>> 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 this finite string pair.
>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>> to take you through a series of student exercises that I am confident
>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>> understanding what this means.
>>>>>
>>>>> OK let's give it a shot.
>>>>
>>>> Great! Let's start with something that will force a lot of hidden
>>>> assumptions to be made explicit.
>>>>
>>>> Q: Is the set of even numbers decidable? If so, specify the TM (call it
>>>> E) using Linz's notation (extended if you like). If not, why not?
>>>>
>>>> (This may be slow to get started, because there is a lot low-level
>>>> things to iron out first.)
>>>>
>>>
>>> That depends on how you define your terms. Any element of the set of
>>> integers can be determined to be divisible by 2 with or without a
>>> remainder.
>>>
>>> Can every element of this set be enumerated in finite time?
>>> No. Can the set of all even numbers be defined in finite space?
>>> Yes through algorithmic compression.
>> I forgot decidability is merely a membership algorithm, so yes.
>>
> Yes. The question is how the input is to be encoded. Whilst you could use
> any encoding, the obvious options are a series of N "1"s followed by a blank,
> or a number in binary, followed by a blank.
> Next thing to do is to write the Turing machine that accepts an even number,
> and rejects everything that isn't an even number. For simplicity, say that all
> inputs will be either odd numbers or even numbers, and any input that isn't a
> number can be disregarded.
> Note that the numbers can be of arbitrary length, though not infinite length.
> That's one of the nice things about Turing machines. Unlike C programs,
> the machines tend to handle huge inputs without any modification.
>
> You'll need some environment for specifying and test-running Turing machines
> to get anywhere very fast. There must be a good online resource.

The key question is it utterly impossible or not? As long as it can be
written as a pure function in C then that proves it is decidable.
I would use ASCII digits as the basis. Simply looking at the last digit
derives the whole algorithm: {0,2,4,6,8}

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<87o81hbx01.fsf@bsb.me.uk>

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 by: Ben Bacarisse - Sun, 3 Apr 2022 22:52 UTC

olcott <NoOne@NoWhere.com> writes:

> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> 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 this finite string pair.
>>>> You certainly have trouble understanding this so I will repeat my offer
>>>> to take you through a series of student exercises that I am confident
>>>> (provided you approach them with an open mind) will lead you to fully
>>>> understanding what this means.
>>>
>>> OK let's give it a shot.
>> Great! Let's start with something that will force a lot of hidden
>> assumptions to be made explicit.
>> Q: Is the set of even numbers decidable? If so, specify the TM (call it
>> E) using Linz's notation (extended if you like). If not, why not?
>> (This may be slow to get started, because there is a lot low-level
>> things to iron out first.)
>
> That depends on how you define your terms. Any element of the set of
> integers can be determined to be divisible by 2 with or without a
> remainder.

So try to define them. What do you think it means for set to be
decidable? What other things do you think need to be defined? Have a
go at defining them, and I'll step in if I disagree. If you have
trouble defining any of these, you can ask for my definition, but you'll
have to accept it!

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<1ce913ac-a7df-49ed-928f-1570ccdd01b0n@googlegroups.com>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [
key missing piece in dialogue ]
From: malcolm....@gmail.com (Malcolm McLean)
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 by: Malcolm McLean - Sun, 3 Apr 2022 22:53 UTC

On Sunday, 3 April 2022 at 23:44:30 UTC+1, olcott wrote:
> On 4/3/2022 5:25 PM, Malcolm McLean wrote:
> > On Sunday, 3 April 2022 at 23:10:26 UTC+1, olcott wrote:
> >> On 4/3/2022 5:07 PM, olcott wrote:
> >>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
> >>>> olcott <No...@NoWhere.com> writes:
> >>>>
> >>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
> >>>>>> olcott <No...@NoWhere.com> writes:
> >>>>>>
> >>>>>>> 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 this finite string pair.
> >>>>>> You certainly have trouble understanding this so I will repeat my offer
> >>>>>> to take you through a series of student exercises that I am confident
> >>>>>> (provided you approach them with an open mind) will lead you to fully
> >>>>>> understanding what this means.
> >>>>>
> >>>>> OK let's give it a shot.
> >>>>
> >>>> Great! Let's start with something that will force a lot of hidden
> >>>> assumptions to be made explicit.
> >>>>
> >>>> Q: Is the set of even numbers decidable? If so, specify the TM (call it
> >>>> E) using Linz's notation (extended if you like). If not, why not?
> >>>>
> >>>> (This may be slow to get started, because there is a lot low-level
> >>>> things to iron out first.)
> >>>>
> >>>
> >>> That depends on how you define your terms. Any element of the set of
> >>> integers can be determined to be divisible by 2 with or without a
> >>> remainder.
> >>>
> >>> Can every element of this set be enumerated in finite time?
> >>> No. Can the set of all even numbers be defined in finite space?
> >>> Yes through algorithmic compression.
> >> I forgot decidability is merely a membership algorithm, so yes.
> >>
> > Yes. The question is how the input is to be encoded. Whilst you could use
> > any encoding, the obvious options are a series of N "1"s followed by a blank,
> > or a number in binary, followed by a blank.
> > Next thing to do is to write the Turing machine that accepts an even number,
> > and rejects everything that isn't an even number. For simplicity, say that all
> > inputs will be either odd numbers or even numbers, and any input that isn't a
> > number can be disregarded.
> > Note that the numbers can be of arbitrary length, though not infinite length.
> > That's one of the nice things about Turing machines. Unlike C programs,
> > the machines tend to handle huge inputs without any modification.
> >
> > You'll need some environment for specifying and test-running Turing machines
> > to get anywhere very fast. There must be a good online resource.
> The key question is it utterly impossible or not? As long as it can be
> written as a pure function in C then that proves it is decidable.
> I would use ASCII digits as the basis. Simply looking at the last digit
> derives the whole algorithm: {0,2,4,6,8}
>
In C it is not possible because you can always create a number too large for
your C program to handle. However for a Turing machine, it ought to be possible.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<87ilrpbwrt.fsf@bsb.me.uk>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
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 by: Ben Bacarisse - Sun, 3 Apr 2022 22:57 UTC

olcott <NoOne@NoWhere.com> writes:

> On 4/3/2022 5:07 PM, olcott wrote:
>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> 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 this finite string pair.
>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>> to take you through a series of student exercises that I am confident
>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>> understanding what this means.
>>>>
>>>> OK let's give it a shot.
>>>
>>> Great!  Let's start with something that will force a lot of hidden
>>> assumptions to be made explicit.
>>>
>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>
>>> (This may be slow to get started, because there is a lot low-level
>>> things to iron out first.)
>>>
>> That depends on how you define your terms. Any element of the set of
>> integers can be determined to be divisible by 2 with or without a remainder.
>> Can every element of this set be enumerated in finite time?
>> No. Can the set of all even numbers be defined in finite space?
>> Yes through algorithmic compression.
>
> I forgot decidability is merely a membership algorithm, so yes.

(Didn't see you'd moved on. Ignore my last reply.)

I would say no. And for the reason you keep giving: any TM that decides
membership can't have a number as input. TMs accept or reject strings,
not numbers.

But surely the iseven(n) function is computable, so how do you think we
should resolve this? Hint: think encoding.

Can you have a stab at specifying E now using Linz's notation?

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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 by: Richard Damon - Sun, 3 Apr 2022 23:23 UTC

On 4/3/22 6:57 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 4/3/2022 5:07 PM, olcott wrote:
>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> 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 this finite string pair.
>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>> to take you through a series of student exercises that I am confident
>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>> understanding what this means.
>>>>>
>>>>> OK let's give it a shot.
>>>>
>>>> Great!  Let's start with something that will force a lot of hidden
>>>> assumptions to be made explicit.
>>>>
>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>
>>>> (This may be slow to get started, because there is a lot low-level
>>>> things to iron out first.)
>>>>
>>> That depends on how you define your terms. Any element of the set of
>>> integers can be determined to be divisible by 2 with or without a remainder.
>>> Can every element of this set be enumerated in finite time?
>>> No. Can the set of all even numbers be defined in finite space?
>>> Yes through algorithmic compression.
>>
>> I forgot decidability is merely a membership algorithm, so yes.
>
> (Didn't see you'd moved on. Ignore my last reply.)
>
> I would say no. And for the reason you keep giving: any TM that decides
> membership can't have a number as input. TMs accept or reject strings,
> not numbers.
>
> But surely the iseven(n) function is computable, so how do you think we
> should resolve this? Hint: think encoding.
>
> Can you have a stab at specifying E now using Linz's notation?
>

And I would say YES, because they can be given a REPRESENTATION of a Number.

Unless the criteria for the Decider is writen based on the string set of
the TM, most deciders work with representations. (And and almost all
interesting functions/decisions aren't based on the specific alphabet of
a given Turing Machine)

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<n72dnae_RMuWrdf_nZ2dnUU7_83NnZ2d@giganews.com>

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key missing piece in dialogue ]
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 by: olcott - Sun, 3 Apr 2022 23:38 UTC

On 4/3/2022 5:57 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 4/3/2022 5:07 PM, olcott wrote:
>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> 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 this finite string pair.
>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>> to take you through a series of student exercises that I am confident
>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>> understanding what this means.
>>>>>
>>>>> OK let's give it a shot.
>>>>
>>>> Great!  Let's start with something that will force a lot of hidden
>>>> assumptions to be made explicit.
>>>>
>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>
>>>> (This may be slow to get started, because there is a lot low-level
>>>> things to iron out first.)
>>>>
>>> That depends on how you define your terms. Any element of the set of
>>> integers can be determined to be divisible by 2 with or without a remainder.
>>> Can every element of this set be enumerated in finite time?
>>> No. Can the set of all even numbers be defined in finite space?
>>> Yes through algorithmic compression.
>>
>> I forgot decidability is merely a membership algorithm, so yes.
>
> (Didn't see you'd moved on. Ignore my last reply.)
>
> I would say no. And for the reason you keep giving: any TM that decides
> membership can't have a number as input. TMs accept or reject strings,
> not numbers.
>
> But surely the iseven(n) function is computable, so how do you think we
> should resolve this? Hint: think encoding.
>
> Can you have a stab at specifying E now using Linz's notation?
>

I want to very thoroughly go through all of the points completely as
efficiently as possible. Because a C function could do this as a pure
function of its ASCII digit sequence inputs by merely examining the last
digits for: {0,2,4,8} we know that E is decidable.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<87czhxbuu0.fsf@bsb.me.uk>

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Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
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 by: Ben Bacarisse - Sun, 3 Apr 2022 23:39 UTC

Richard Damon <Richard@Damon-Family.org> writes:

> On 4/3/22 6:57 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> 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 this finite string pair.
>>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>>> to take you through a series of student exercises that I am confident
>>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>>> understanding what this means.
>>>>>>
>>>>>> OK let's give it a shot.
>>>>>
>>>>> Great!  Let's start with something that will force a lot of hidden
>>>>> assumptions to be made explicit.
>>>>>
>>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>>
>>>>> (This may be slow to get started, because there is a lot low-level
>>>>> things to iron out first.)
>>>>>
>>>> That depends on how you define your terms. Any element of the set of
>>>> integers can be determined to be divisible by 2 with or without a remainder.
>>>> Can every element of this set be enumerated in finite time?
>>>> No. Can the set of all even numbers be defined in finite space?
>>>> Yes through algorithmic compression.
>>>
>>> I forgot decidability is merely a membership algorithm, so yes.
>> (Didn't see you'd moved on. Ignore my last reply.)
>> I would say no. And for the reason you keep giving: any TM that decides
>> membership can't have a number as input. TMs accept or reject strings,
>> not numbers.
>> But surely the iseven(n) function is computable, so how do you think we
>> should resolve this? Hint: think encoding.
>> Can you have a stab at specifying E now using Linz's notation?
>>
>
> And I would say YES, because they can be given a REPRESENTATION of a
> Number.

And you are welcome to say that. You won't be confused by saying it.

The trouble with a blanket yes is what representations can we use?
There are representations of TMs that make halting decidable but, of
course, they are not computable from the obvious ones. I want to stress
to PO that some agreed representation is the very starting point of a
decision problem.

If this were 17 years ago I would not allow PO to use any old "encoding"
for TMs. Everything would be unary representations of numbers. The
only decidable sets would be subsets of N, determined by the
accept/reject of the unary encoding of the number. The halting problem
would be express in terms of TM number i with input number j. But that
ship sailed a long time ago.

> Unless the criteria for the Decider is writen based on the string set
> of the TM, most deciders work with representations. (And and almost
> all interesting functions/decisions aren't based on the specific
> alphabet of a given Turing Machine)

Too abstract, I think, for PO. Also, I want him to write a TM and that
will need a specific alphabet.

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<n72dnaa_RMs0rdf_nZ2dnUU7_81QAAAA@giganews.com>

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 by: olcott - Sun, 3 Apr 2022 23:41 UTC

On 4/3/2022 5:53 PM, Malcolm McLean wrote:
> On Sunday, 3 April 2022 at 23:44:30 UTC+1, olcott wrote:
>> On 4/3/2022 5:25 PM, Malcolm McLean wrote:
>>> On Sunday, 3 April 2022 at 23:10:26 UTC+1, olcott wrote:
>>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>
>>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> 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 this finite string pair.
>>>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>>>> to take you through a series of student exercises that I am confident
>>>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>>>> understanding what this means.
>>>>>>>
>>>>>>> OK let's give it a shot.
>>>>>>
>>>>>> Great! Let's start with something that will force a lot of hidden
>>>>>> assumptions to be made explicit.
>>>>>>
>>>>>> Q: Is the set of even numbers decidable? If so, specify the TM (call it
>>>>>> E) using Linz's notation (extended if you like). If not, why not?
>>>>>>
>>>>>> (This may be slow to get started, because there is a lot low-level
>>>>>> things to iron out first.)
>>>>>>
>>>>>
>>>>> That depends on how you define your terms. Any element of the set of
>>>>> integers can be determined to be divisible by 2 with or without a
>>>>> remainder.
>>>>>
>>>>> Can every element of this set be enumerated in finite time?
>>>>> No. Can the set of all even numbers be defined in finite space?
>>>>> Yes through algorithmic compression.
>>>> I forgot decidability is merely a membership algorithm, so yes.
>>>>
>>> Yes. The question is how the input is to be encoded. Whilst you could use
>>> any encoding, the obvious options are a series of N "1"s followed by a blank,
>>> or a number in binary, followed by a blank.
>>> Next thing to do is to write the Turing machine that accepts an even number,
>>> and rejects everything that isn't an even number. For simplicity, say that all
>>> inputs will be either odd numbers or even numbers, and any input that isn't a
>>> number can be disregarded.
>>> Note that the numbers can be of arbitrary length, though not infinite length.
>>> That's one of the nice things about Turing machines. Unlike C programs,
>>> the machines tend to handle huge inputs without any modification.
>>>
>>> You'll need some environment for specifying and test-running Turing machines
>>> to get anywhere very fast. There must be a good online resource.
>> The key question is it utterly impossible or not? As long as it can be
>> written as a pure function in C then that proves it is decidable.
>> I would use ASCII digits as the basis. Simply looking at the last digit
>> derives the whole algorithm: {0,2,4,6,8}
>>
> In C it is not possible because you can always create a number too large for
> your C program to handle. However for a Turing machine, it ought to be possible.
>

I will never tolerate the tediousness of encoding TMs, my proof is
sufficient.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<n72dnaG_RMvdrNf_nZ2dnUU7_81g4p2d@giganews.com>

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 by: olcott - Sun, 3 Apr 2022 23:43 UTC

On 4/3/2022 6:39 PM, Ben Bacarisse wrote:
> Richard Damon <Richard@Damon-Family.org> writes:
>
>> On 4/3/22 6:57 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> 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 this finite string pair.
>>>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>>>> to take you through a series of student exercises that I am confident
>>>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>>>> understanding what this means.
>>>>>>>
>>>>>>> OK let's give it a shot.
>>>>>>
>>>>>> Great!  Let's start with something that will force a lot of hidden
>>>>>> assumptions to be made explicit.
>>>>>>
>>>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>>>
>>>>>> (This may be slow to get started, because there is a lot low-level
>>>>>> things to iron out first.)
>>>>>>
>>>>> That depends on how you define your terms. Any element of the set of
>>>>> integers can be determined to be divisible by 2 with or without a remainder.
>>>>> Can every element of this set be enumerated in finite time?
>>>>> No. Can the set of all even numbers be defined in finite space?
>>>>> Yes through algorithmic compression.
>>>>
>>>> I forgot decidability is merely a membership algorithm, so yes.
>>> (Didn't see you'd moved on. Ignore my last reply.)
>>> I would say no. And for the reason you keep giving: any TM that decides
>>> membership can't have a number as input. TMs accept or reject strings,
>>> not numbers.
>>> But surely the iseven(n) function is computable, so how do you think we
>>> should resolve this? Hint: think encoding.
>>> Can you have a stab at specifying E now using Linz's notation?
>>>
>>
>> And I would say YES, because they can be given a REPRESENTATION of a
>> Number.
>
> And you are welcome to say that. You won't be confused by saying it.
>
> The trouble with a blanket yes is what representations can we use?

All of this is totally extraneous.
Is a memebership algorithm for E possible? YES any C program could
trivially implement it.

> There are representations of TMs that make halting decidable but, of
> course, they are not computable from the obvious ones. I want to stress
> to PO that some agreed representation is the very starting point of a
> decision problem.
>
> If this were 17 years ago I would not allow PO to use any old "encoding"
> for TMs. Everything would be unary representations of numbers. The
> only decidable sets would be subsets of N, determined by the
> accept/reject of the unary encoding of the number. The halting problem
> would be express in terms of TM number i with input number j. But that
> ship sailed a long time ago.
>
>> Unless the criteria for the Decider is writen based on the string set
>> of the TM, most deciders work with representations. (And and almost
>> all interesting functions/decisions aren't based on the specific
>> alphabet of a given Turing Machine)
>
> Too abstract, I think, for PO. Also, I want him to write a TM and that
> will need a specific alphabet.
>

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<877d85buic.fsf@bsb.me.uk>

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Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
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 by: Ben Bacarisse - Sun, 3 Apr 2022 23:46 UTC

olcott <NoOne@NoWhere.com> writes:

> On 4/3/2022 5:57 PM, Ben Bacarisse wrote:
>> olcott <NoOne@NoWhere.com> writes:
>>
>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> 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 this finite string pair.
>>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>>> to take you through a series of student exercises that I am confident
>>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>>> understanding what this means.
>>>>>>
>>>>>> OK let's give it a shot.
>>>>>
>>>>> Great!  Let's start with something that will force a lot of hidden
>>>>> assumptions to be made explicit.
>>>>>
>>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>>
>>>>> (This may be slow to get started, because there is a lot low-level
>>>>> things to iron out first.)
>>>>>
>>>> That depends on how you define your terms. Any element of the set of
>>>> integers can be determined to be divisible by 2 with or without a remainder.
>>>> Can every element of this set be enumerated in finite time?
>>>> No. Can the set of all even numbers be defined in finite space?
>>>> Yes through algorithmic compression.
>>>
>>> I forgot decidability is merely a membership algorithm, so yes.
>>
>> (Didn't see you'd moved on. Ignore my last reply.)
>> I would say no. And for the reason you keep giving: any TM that decides
>> membership can't have a number as input. TMs accept or reject strings,
>> not numbers.
>>
>> But surely the iseven(n) function is computable, so how do you think we
>> should resolve this? Hint: think encoding.
>> Can you have a stab at specifying E now using Linz's notation?
>
> I want to very thoroughly go through all of the points completely as
> efficiently as possible. Because a C function could do this as a pure
> function of its ASCII digit sequence inputs by merely examining the
> last digits for: {0,2,4,8} we know that E is decidable.

It will take longer if you take digressions like this! C is not a
complete model of computation. No C function can do the job for any
number.

Let's stick to TMs. What encoding would you use for a TM decider for
this problem? Can you specify it using Linz's notation?

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<871qydbu46.fsf@bsb.me.uk>

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Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
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 by: Ben Bacarisse - Sun, 3 Apr 2022 23:54 UTC

olcott <NoOne@NoWhere.com> writes:

> On 4/3/2022 6:39 PM, Ben Bacarisse wrote:
>> Richard Damon <Richard@Damon-Family.org> writes:
>>
>>> On 4/3/22 6:57 PM, Ben Bacarisse wrote:
>>>> olcott <NoOne@NoWhere.com> writes:
>>>>
>>>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> 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 this finite string pair.
>>>>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>>>>> to take you through a series of student exercises that I am confident
>>>>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>>>>> understanding what this means.
>>>>>>>>
>>>>>>>> OK let's give it a shot.
>>>>>>>
>>>>>>> Great!  Let's start with something that will force a lot of hidden
>>>>>>> assumptions to be made explicit.
>>>>>>>
>>>>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>>>>
>>>>>>> (This may be slow to get started, because there is a lot low-level
>>>>>>> things to iron out first.)
>>>>>>>
>>>>>> That depends on how you define your terms. Any element of the set of
>>>>>> integers can be determined to be divisible by 2 with or without a remainder.
>>>>>> Can every element of this set be enumerated in finite time?
>>>>>> No. Can the set of all even numbers be defined in finite space?
>>>>>> Yes through algorithmic compression.
>>>>>
>>>>> I forgot decidability is merely a membership algorithm, so yes.
>>>> (Didn't see you'd moved on. Ignore my last reply.)
>>>> I would say no. And for the reason you keep giving: any TM that decides
>>>> membership can't have a number as input. TMs accept or reject strings,
>>>> not numbers.
>>>> But surely the iseven(n) function is computable, so how do you think we
>>>> should resolve this? Hint: think encoding.
>>>> Can you have a stab at specifying E now using Linz's notation?
>>>>
>>>
>>> And I would say YES, because they can be given a REPRESENTATION of a
>>> Number.
>> And you are welcome to say that. You won't be confused by saying it.
>> The trouble with a blanket yes is what representations can we use?
>
> All of this is totally extraneous.
> Is a memebership algorithm for E possible? YES any C program could
> trivially implement it.

You are not in "student mode" here. You are laying down the law, but
that's my job!

First, E is the name I suggested a TM, not for the set of even numbers.
You must use the names I suggest for the things I suggest you do or
there will be dozens of posts just about what E is or what X is!

Second, you don't yet know enough about compatibility theory to switch
between TMs and C. C is not suitable for lots of reasons.

Let's stick to TMs until that is fully wrapped up.

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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Newsgroups: comp.theory
Subject: Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]
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 by: Ben Bacarisse - Sun, 3 Apr 2022 23:58 UTC

olcott <NoOne@NoWhere.com> writes:

> I will never tolerate the tediousness of encoding TMs, my proof is
> sufficient.

Oh! I thought you were entering into this with an open mind and that
you were prepared to do it properly. Encoding TMs is central to the
halting problem.

Do you want to stop?

--
Ben.

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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 by: olcott - Mon, 4 Apr 2022 00:01 UTC

On 4/3/2022 6:46 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 4/3/2022 5:57 PM, Ben Bacarisse wrote:
>>> olcott <NoOne@NoWhere.com> writes:
>>>
>>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>
>>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> 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 this finite string pair.
>>>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>>>> to take you through a series of student exercises that I am confident
>>>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>>>> understanding what this means.
>>>>>>>
>>>>>>> OK let's give it a shot.
>>>>>>
>>>>>> Great!  Let's start with something that will force a lot of hidden
>>>>>> assumptions to be made explicit.
>>>>>>
>>>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>>>
>>>>>> (This may be slow to get started, because there is a lot low-level
>>>>>> things to iron out first.)
>>>>>>
>>>>> That depends on how you define your terms. Any element of the set of
>>>>> integers can be determined to be divisible by 2 with or without a remainder.
>>>>> Can every element of this set be enumerated in finite time?
>>>>> No. Can the set of all even numbers be defined in finite space?
>>>>> Yes through algorithmic compression.
>>>>
>>>> I forgot decidability is merely a membership algorithm, so yes.
>>>
>>> (Didn't see you'd moved on. Ignore my last reply.)
>>> I would say no. And for the reason you keep giving: any TM that decides
>>> membership can't have a number as input. TMs accept or reject strings,
>>> not numbers.
>>>
>>> But surely the iseven(n) function is computable, so how do you think we
>>> should resolve this? Hint: think encoding.
>>> Can you have a stab at specifying E now using Linz's notation?
>>
>> I want to very thoroughly go through all of the points completely as
>> efficiently as possible. Because a C function could do this as a pure
>> function of its ASCII digit sequence inputs by merely examining the
>> last digits for: {0,2,4,8} we know that E is decidable.
>
> It will take longer if you take digressions like this! C is not a
> complete model of computation. No C function can do the job for any
> number.
>
> Let's stick to TMs. What encoding would you use for a TM decider for
> this problem? Can you specify it using Linz's notation?
>

ASCII text digits. No.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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 by: olcott - Mon, 4 Apr 2022 00:03 UTC

On 4/3/2022 6:54 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> On 4/3/2022 6:39 PM, Ben Bacarisse wrote:
>>> Richard Damon <Richard@Damon-Family.org> writes:
>>>
>>>> On 4/3/22 6:57 PM, Ben Bacarisse wrote:
>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>
>>>>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>
>>>>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>>>>> olcott <NoOne@NoWhere.com> writes:
>>>>>>>>>>
>>>>>>>>>>> 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 this finite string pair.
>>>>>>>>>> You certainly have trouble understanding this so I will repeat my offer
>>>>>>>>>> to take you through a series of student exercises that I am confident
>>>>>>>>>> (provided you approach them with an open mind) will lead you to fully
>>>>>>>>>> understanding what this means.
>>>>>>>>>
>>>>>>>>> OK let's give it a shot.
>>>>>>>>
>>>>>>>> Great!  Let's start with something that will force a lot of hidden
>>>>>>>> assumptions to be made explicit.
>>>>>>>>
>>>>>>>> Q: Is the set of even numbers decidable?  If so, specify the TM (call it
>>>>>>>>     E) using Linz's notation (extended if you like).  If not, why not?
>>>>>>>>
>>>>>>>> (This may be slow to get started, because there is a lot low-level
>>>>>>>> things to iron out first.)
>>>>>>>>
>>>>>>> That depends on how you define your terms. Any element of the set of
>>>>>>> integers can be determined to be divisible by 2 with or without a remainder.
>>>>>>> Can every element of this set be enumerated in finite time?
>>>>>>> No. Can the set of all even numbers be defined in finite space?
>>>>>>> Yes through algorithmic compression.
>>>>>>
>>>>>> I forgot decidability is merely a membership algorithm, so yes.
>>>>> (Didn't see you'd moved on. Ignore my last reply.)
>>>>> I would say no. And for the reason you keep giving: any TM that decides
>>>>> membership can't have a number as input. TMs accept or reject strings,
>>>>> not numbers.
>>>>> But surely the iseven(n) function is computable, so how do you think we
>>>>> should resolve this? Hint: think encoding.
>>>>> Can you have a stab at specifying E now using Linz's notation?
>>>>>
>>>>
>>>> And I would say YES, because they can be given a REPRESENTATION of a
>>>> Number.
>>> And you are welcome to say that. You won't be confused by saying it.
>>> The trouble with a blanket yes is what representations can we use?
>>
>> All of this is totally extraneous.
>> Is a memebership algorithm for E possible? YES any C program could
>> trivially implement it.
>
> You are not in "student mode" here. You are laying down the law, but
> that's my job!
>
> First, E is the name I suggested a TM, not for the set of even numbers.
> You must use the names I suggest for the things I suggest you do or
> there will be dozens of posts just about what E is or what X is!
>
> Second, you don't yet know enough about compatibility theory to switch
> between TMs and C. C is not suitable for lots of reasons.
>
> Let's stick to TMs until that is fully wrapped up.
>

Any encoding besides ASCII text digits is intolerable.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

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 by: olcott - Mon, 4 Apr 2022 00:04 UTC

On 4/3/2022 6:58 PM, Ben Bacarisse wrote:
> olcott <NoOne@NoWhere.com> writes:
>
>> I will never tolerate the tediousness of encoding TMs, my proof is
>> sufficient.
>
> Oh! I thought you were entering into this with an open mind and that
> you were prepared to do it properly. Encoding TMs is central to the
> halting problem.
>
> Do you want to stop?
>

I will not tolerate TMs that cannot store an ASCII digit as a single
tape element.

--
Copyright 2022 Pete Olcott

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

Re: Refuting the Peter Linz Halting Problem Proof --- Version(10) [ key missing piece in dialogue ]

<LEq2K.501518$Rza5.173859@fx47.iad>

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 by: Richard Damon - Mon, 4 Apr 2022 00:07 UTC

On 4/3/22 7:41 PM, olcott wrote:
> On 4/3/2022 5:53 PM, Malcolm McLean wrote:
>> On Sunday, 3 April 2022 at 23:44:30 UTC+1, olcott wrote:
>>> On 4/3/2022 5:25 PM, Malcolm McLean wrote:
>>>> On Sunday, 3 April 2022 at 23:10:26 UTC+1, olcott wrote:
>>>>> On 4/3/2022 5:07 PM, olcott wrote:
>>>>>> On 4/3/2022 4:56 PM, Ben Bacarisse wrote:
>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>
>>>>>>>> On 4/3/2022 2:02 PM, Ben Bacarisse wrote:
>>>>>>>>> olcott <No...@NoWhere.com> writes:
>>>>>>>>>
>>>>>>>>>> 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 this finite string pair.
>>>>>>>>> You certainly have trouble understanding this so I will repeat
>>>>>>>>> my offer
>>>>>>>>> to take you through a series of student exercises that I am
>>>>>>>>> confident
>>>>>>>>> (provided you approach them with an open mind) will lead you to
>>>>>>>>> fully
>>>>>>>>> understanding what this means.
>>>>>>>>
>>>>>>>> OK let's give it a shot.
>>>>>>>
>>>>>>> Great! Let's start with something that will force a lot of hidden
>>>>>>> assumptions to be made explicit.
>>>>>>>
>>>>>>> Q: Is the set of even numbers decidable? If so, specify the TM
>>>>>>> (call it
>>>>>>> E) using Linz's notation (extended if you like). If not, why not?
>>>>>>>
>>>>>>> (This may be slow to get started, because there is a lot low-level
>>>>>>> things to iron out first.)
>>>>>>>
>>>>>>
>>>>>> That depends on how you define your terms. Any element of the set of
>>>>>> integers can be determined to be divisible by 2 with or without a
>>>>>> remainder.
>>>>>>
>>>>>> Can every element of this set be enumerated in finite time?
>>>>>> No. Can the set of all even numbers be defined in finite space?
>>>>>> Yes through algorithmic compression.
>>>>> I forgot decidability is merely a membership algorithm, so yes.
>>>>>
>>>> Yes. The question is how the input is to be encoded. Whilst you
>>>> could use
>>>> any encoding, the obvious options are a series of N "1"s followed by
>>>> a blank,
>>>> or a number in binary, followed by a blank.
>>>> Next thing to do is to write the Turing machine that accepts an even
>>>> number,
>>>> and rejects everything that isn't an even number. For simplicity,
>>>> say that all
>>>> inputs will be either odd numbers or even numbers, and any input
>>>> that isn't a
>>>> number can be disregarded.
>>>> Note that the numbers can be of arbitrary length, though not
>>>> infinite length.
>>>> That's one of the nice things about Turing machines. Unlike C programs,
>>>> the machines tend to handle huge inputs without any modification.
>>>>
>>>> You'll need some environment for specifying and test-running Turing
>>>> machines
>>>> to get anywhere very fast. There must be a good online resource.
>>> The key question is it utterly impossible or not? As long as it can be
>>> written as a pure function in C then that proves it is decidable.
>>> I would use ASCII digits as the basis. Simply looking at the last digit
>>> derives the whole algorithm: {0,2,4,6,8}
>>>
>> In C it is not possible because you can always create a number too
>> large for
>> your C program to handle. However for a Turing machine, it ought to be
>> possible.
>>
>
> I will never tolerate the tediousness of encoding TMs, my proof is
> sufficient.
>

Your proof is in ERROR because you don't understand what a TM is.

You make claims about what a TM could do that you can not actually prove.

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server_pubkey.txt

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