Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.

/Flibble

--- SoupGate-Win32 v1.05
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On 2025-05-15 06:27:13 +0000, Mr Flibble said:

Peter is right to say that the halting problem as defined is flawed: he agrees with me that there is category error at the heart of the problem definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

No, he is not right about that. There is no flaw about the problem. The
problem is to create a halt decider. Every Turing machine either is or
is not a halt decider. In order to demonstrate that a Turing machine is
not a halt decider it is sufficient to show one example that it does
not determine correctly.

That a problem is too hard to you does not mean that it be ill-posed.

--
Mikko

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 5/15/25 2:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he agrees with me that there is category error at the heart of the problem definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.

/Flibble

There is nothing wrong with the Halting Problem as actually defined.

There is a lot wrong with how he interprests the Problem, because he
doesn't understand the meaning of the basic terms.

First, H needs to be a fully defined program, and its input the
representatio of another fully defined program.

When he talks about an infinite set of them pairs, they each are totally different problems, as the input is diffferent for each case, and thus
his "induciton" isn't valid. That fact that none of his versions of H
can correctly get to the end of their version of D is meaningless, and
doesn't show that it is non-halting, when the truth is that for every H
that does abort and return 0, the actual behavior of the program its
input represents (which calls that H that returns 0) is to halt.

Yes, the Hs that never abort and return 0 will create a D that will
never return, but those are different inputs than above, and none of
those Hs ever give an answer and thus fail to be the decider asked for.

So, Peter's attempt to refute the problem proof, just recreates a step
of the proof showing that no H created by his method ever gvies the
right answer, CONFIRMING (not refuting) the proof.

The only world where Peter is correct, would be in some alternate POOPS
theory with his totally strange set of rules, which is why no one will
every want to use his POOPS if he ever does try to fomralize it.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Thu, 15 May 2025 07:27:03 -0400, Richard Damon wrote:

On 5/15/25 2:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is
equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.

/Flibble

There is nothing wrong with the Halting Problem as actually defined.

There is a lot wrong with how he interprests the Problem, because he
doesn't understand the meaning of the basic terms.

False.

[snip]

The only world where Peter is correct, would be in some alternate POOPS theory with his totally strange set of rules, which is why no one will
every want to use his POOPS if he ever does try to fomralize it.

Grow up.

/Flibble

--- SoupGate-Win32 v1.05
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Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

that part Turing et al got right.

Turing never said that there are undecidable inputs[2].

Maybe "truth", "pathological", "input" and "undecidable" have special
Flibble meanings. I'm willing to accept that "the" and "is" have the
usual semantics.

[1] By input I mean an instance of the halting problem -- a string of
symbols representing (a) an encoded TM (a number is Turing's paper)
and (b) the initial tape contents.

[2] In the original paper, he never uses the words "input" or
"decidable". Instead, he uses other words, but nowhere is there any
remark that is even close to meaning what you say.

--
Ben.

--- SoupGate-Win32 v1.05
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On Thu, 15 May 2025 11:07:48 +0300, Mikko wrote:

On 2025-05-15 06:27:13 +0000, Mr Flibble said:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

No, he is not right about that. There is no flaw about the problem. The problem is to create a halt decider. Every Turing machine either is or
is not a halt decider. In order to demonstrate that a Turing machine is
not a halt decider it is sufficient to show one example that it does not determine correctly.

False -- the pathologial input cannot be determined correctly because it
is ill-posed.

That a problem is too hard to you does not mean that it be ill-posed.

It isn't the case that problem is too hard to me, it is the case that it
is ill-posed.

/Flibble

--- SoupGate-Win32 v1.05
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On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh? Partial deciders are a thing.

that part Turing et al got right.

Turing never said that there are undecidable inputs[2].

Maybe "truth", "pathological", "input" and "undecidable" have special
Flibble meanings. I'm willing to accept that "the" and "is" have the
usual semantics.

[1] By input I mean an instance of the halting problem -- a string of
symbols representing (a) an encoded TM (a number is Turing's paper)
and (b) the initial tape contents.

[2] In the original paper, he never uses the words "input" or
"decidable". Instead, he uses other words, but nowhere is there any
remark that is even close to meaning what you say.

False.

/Flibble

--- SoupGate-Win32 v1.05
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On Thu, 15 May 2025 10:13:50 -0500, olcott wrote:

On 5/15/2025 1:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is
equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.

/Flibble

Introduction to the Theory of Computation 3rd Edition by Michael Sipser (Author)
4.4 out of 5 stars 568 ratings

https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/

113318779X

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D until
H correctly determines that its simulated D would never stop
running unless aborted then

H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

HHH does correctly reject DDD and DD according to the exact meaning of
the above words. It also seems to me that those words are a truism.

Sipser is wrong: he is disagreeing with Turing et al that pathological
input is undecidable.

/Flibble

--- SoupGate-Win32 v1.05
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On 5/15/25 9:03 AM, Mr Flibble wrote:

On Thu, 15 May 2025 11:07:48 +0300, Mikko wrote:

On 2025-05-15 06:27:13 +0000, Mr Flibble said:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

No, he is not right about that. There is no flaw about the problem. The
problem is to create a halt decider. Every Turing machine either is or
is not a halt decider. In order to demonstrate that a Turing machine is
not a halt decider it is sufficient to show one example that it does not
determine correctly.

False -- the pathologial input cannot be determined correctly because it
is ill-posed.

What is "ill-posed" about it. Remember, The decide is a specific and
determined deterministic program, and thus we can compute the answer it
WILL give for any input. The "pathological" is also precisely built on
that decider, and also is fully deterministic. It begins by using its
copy of the algorithm of the decider to get the answer it will give for
itself (in the standard proof, from the representation of itself given
to it as its input) and then do the opposite.

Which action of that is impossible?

If the decider doesn't process that particular string, it just failed to
be4 the required decider. If it does, then the pathological program can
just act on the answer it got.

That a problem is too hard to you does not mean that it be ill-posed.

It isn't the case that problem is too hard to me, it is the case that it
is ill-posed.

/Flibble

Nope, just your brain is ill-processing.

--- SoupGate-Win32 v1.05
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Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh? Partial deciders are a thing.

Yes. That does not alter the fact that no input is undecidable.

that part Turing et al got right.

Turing never said that there are undecidable inputs[2].

Maybe "truth", "pathological", "input" and "undecidable" have special
Flibble meanings. I'm willing to accept that "the" and "is" have the
usual semantics.

[1] By input I mean an instance of the halting problem -- a string of
symbols representing (a) an encoded TM (a number is Turing's paper)
and (b) the initial tape contents.

[2] In the original paper, he never uses the words "input" or
"decidable". Instead, he uses other words, but nowhere is there any
remark that is even close to meaning what you say.

False.

Not an argument one can counter. Well done!

--
Ben.

--- SoupGate-Win32 v1.05
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On Fri, 16 May 2025 00:59:02 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh? Partial deciders are a thing.

Yes. That does not alter the fact that no input is undecidable.

Pathological input is undecidable as pathological input is an "impossible program" [Strachey 1965].

/Flibble

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 16/05/2025 00:59, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh? Partial deciders are a thing.

Yes. That does not alter the fact that no input is undecidable.

that part Turing et al got right.

Turing never said that there are undecidable inputs[2].

Maybe "truth", "pathological", "input" and "undecidable" have special
Flibble meanings. I'm willing to accept that "the" and "is" have the
usual semantics.

[1] By input I mean an instance of the halting problem -- a string of
symbols representing (a) an encoded TM (a number is Turing's paper) >>> and (b) the initial tape contents.

[2] In the original paper, he never uses the words "input" or
"decidable". Instead, he uses other words, but nowhere is there any >>> remark that is even close to meaning what you say.

False.

Not an argument one can counter. Well done!

I'll take a crack at it, if I may.

True.

That is: In the original paper, he never uses the words "input"
or "decidable".

(He doesn't even use the word 'halt'.)

--
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within

--- SoupGate-Win32 v1.05
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On 2025-05-15 13:01:25 +0000, Mr Flibble said:

On Thu, 15 May 2025 07:27:03 -0400, Richard Damon wrote:

On 5/15/25 2:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is
equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.

/Flibble

There is nothing wrong with the Halting Problem as actually defined.

There is a lot wrong with how he interprests the Problem, because he
doesn't understand the meaning of the basic terms.

False.

If it were false you could post corrections to commets that identify
a wrong interpretation of the problem or a failure to understand the
meaning of some basic term.

--
Mikko

--- SoupGate-Win32 v1.05
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On 2025-05-15 13:03:19 +0000, Mr Flibble said:

On Thu, 15 May 2025 11:07:48 +0300, Mikko wrote:

On 2025-05-15 06:27:13 +0000, Mr Flibble said:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

No, he is not right about that. There is no flaw about the problem. The
problem is to create a halt decider. Every Turing machine either is or
is not a halt decider. In order to demonstrate that a Turing machine is
not a halt decider it is sufficient to show one example that it does not
determine correctly.

False -- the pathologial input cannot be determined correctly because it
is ill-posed.

Olcott's "pathological" inputa are not "ill-posed". They are well defined programs that can be run, and have been, and they halt when run.

--
Mikko

--- SoupGate-Win32 v1.05
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On 2025-05-15 15:33:01 +0000, Mr Flibble said:

On Thu, 15 May 2025 10:13:50 -0500, olcott wrote:

On 5/15/2025 1:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is
equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.

/Flibble

Introduction to the Theory of Computation 3rd Edition by Michael Sipser
(Author)
4.4 out of 5 stars 568 ratings

https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/

113318779X

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D until
H correctly determines that its simulated D would never stop
running unless aborted then

H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

HHH does correctly reject DDD and DD according to the exact meaning of
the above words. It also seems to me that those words are a truism.

Sipser is wrong: he is disagreeing with Turing et al that pathological
input is undecidable.

Which sentence of Sipser contradicts which sentence of Turing?
Why do you think that Sipser is wrong and not Turing?

--
Mikko

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 2025-05-15 15:13:50 +0000, olcott said:

On 5/15/2025 1:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he
agrees with me that there is category error at the heart of the problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is
equivalant to a halting state of non-halting: the truth is pathlogical
input is undecidable: that part Turing et al got right.

/Flibble

Introduction to the Theory of Computation 3rd Edition
by Michael Sipser (Author)
4.4 out of 5 stars 568 ratings

https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X

Nothing on that page supports any of your claims in any way.

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its
input D until H correctly determines that its simulated D
would never stop running unless aborted then

H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

Nothing above supports any of your claims.

HHH does correctly reject DDD and DD according
to the exact meaning of the above words. It also
seems to me that those words are a truism.

HHH does indeed reject DDD and DD but the use of the word "correctly"
is not justified. HHH does not correctly determine that its
simulated D would never stop running unless aborted.

--
Mikko

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Op 16.mei.2025 om 17:34 schreef olcott:

On 5/16/2025 1:54 AM, Mikko wrote:

On 2025-05-15 15:33:01 +0000, Mr Flibble said:

On Thu, 15 May 2025 10:13:50 -0500, olcott wrote:

On 5/15/2025 1:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is
flawed: he
agrees with me that there is category error at the heart of the
problem
definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that
manifests in
his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is >>>>> equivalant to a halting state of non-halting: the truth is pathlogical >>>>> input is undecidable: that part Turing et al got right.

/Flibble

Introduction to the Theory of Computation 3rd Edition by Michael Sipser >>>> (Author)
4.4 out of 5 stars    568 ratings

https://www.amazon.com/Introduction-Theory-Computation-Michael-
Sipser/dp/

113318779X

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D until
H correctly determines that its simulated D would never stop
running unless aborted then

H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022> >>>>
HHH does correctly reject DDD and DD according to the exact meaning of >>>> the above words. It also seems to me that those words are a truism.

Sipser is wrong: he is disagreeing with Turing et al that pathological
input is undecidable.

Which sentence of Sipser contradicts which sentence of Turing?
Why do you think that Sipser is wrong and not Turing?

A simulating halt decider (SHD) (as defined above)
proves that there cannot be any input that actually
does the opposite of whatever value that its SHD
returns.

int main()
{
  DDD(); // The HHH called by DDD() cannot report on its caller.
}

No, it proves that a correct simulation is impossible. Either the
simulation does not halt, or it aborts before it can see that the input
halts. The input specifies a halting program, but the programmer by
programming the abort, the programmer made HHH blind for this
specification. That HHH does skip this important part of the
specification, does not mean that the specification is not present in
the input.
Of course, you will ignore this verifiable facts and just repeat your
claims. Is that because you don't understand them, or because your
dreams overrule the facts? Come out of rebuttal mode and try to think.
Learning something new is not so dangerous as you seem to think.

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

Op 16.mei.2025 om 17:30 schreef olcott:

On 5/16/2025 1:52 AM, Mikko wrote:

On 2025-05-15 15:13:50 +0000, olcott said:

On 5/15/2025 1:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he >>>> agrees with me that there is category error at the heart of the problem >>>> definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in >>>> his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is
equivalant to a halting state of non-halting: the truth is pathlogical >>>> input is undecidable: that part Turing et al got right.

/Flibble

Introduction to the Theory of Computation 3rd Edition
by Michael Sipser (Author)
4.4 out of 5 stars    568 ratings

https://www.amazon.com/Introduction-Theory-Computation-Michael-
Sipser/ dp/113318779X

Nothing on that page supports any of your claims in any way.

*It establishes Professor Sipser as a qualified authority*
(an appeal to a qualified authority is not an inductive error)

<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its
     input D until H correctly determines that its simulated D
     would never stop running unless aborted then

     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

Nothing above supports any of your claims.

Mike explains all of the details of how the
above quote does derive a correct Simulating Halt Decider.

On 5/14/2025 7:36 PM, Mike Terry wrote:

There is a natural (and correct) statement that Sipser
is far more likely (I'd say) to have agreed to.

First you should understand the basic idea behind a
"Simulating Halt Decider" (*SHD*) that /partially/
simulates its input, while observing each simulation
step looking for certain halting/non-halting patterns
in the simulation. A simple (working) example here
is an input which goes into a tight loop.

(Mike says much more about this)

*Click here to get the whole article*
https://al.howardknight.net/? STYPE=msgid&MSGI=%3C1003cu5%242p3g1%241%40dont-email.me%3E

Message-ID: <1003cu5$2p3g1$1@dont-email.me>

HHH does correctly reject DDD and DD according
to the exact meaning of the above words. It also
seems to me that those words are a truism.

HHH does indeed reject DDD and DD but the use of the word "correctly"
is not justified. HHH does not correctly determine that its
simulated D would never stop running unless aborted.

It easier to see this with DDD;

void DDD()
{
  HHH(DDD);
  return;
}

HHH simulates DDD
The simulated DDD calls HHH(DDD)
The simulated HHH simulates DDD
The simulated DDD calls HHH(DDD)
The simulated HHH simulates DDD
The simulated DDD calls HHH(DDD)
The simulated HHH simulates DDD
The simulated DDD calls HHH(DDD)

Counterfactual. The HHH you published aborts after one cycle.
If this input is not changed, the end of the simulation would be reached
after two cycles.
But you confused yourself, by changing the input when you change your
HHH. Changing the input is not allowed.

How many recursive simulations are needed
before you get the idea that DDD correctly
simulated by HHH

*would never stop running unless aborted*

It is aborted, so it does stop.

--- SoupGate-Win32 v1.05
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Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Fri, 16 May 2025 00:59:02 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh? Partial deciders are a thing.

Yes. That does not alter the fact that no input is undecidable.

Pathological input is undecidable as pathological input is an "impossible program" [Strachey 1965].

The most likely explanation is that you don't know what decidable means.
Either that or you just like posting remarks for the sake of it.

--
Ben.

--- SoupGate-Win32 v1.05
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Richard Heathfield <rjh@cpax.org.uk> writes:

On 16/05/2025 00:59, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh? Partial deciders are a thing.

Yes. That does not alter the fact that no input is undecidable.

that part Turing et al got right.

Turing never said that there are undecidable inputs[2].

Maybe "truth", "pathological", "input" and "undecidable" have special
Flibble meanings. I'm willing to accept that "the" and "is" have the
usual semantics.

[1] By input I mean an instance of the halting problem -- a string of
symbols representing (a) an encoded TM (a number is Turing's paper) >>>> and (b) the initial tape contents.

[2] In the original paper, he never uses the words "input" or
"decidable". Instead, he uses other words, but nowhere is there any >>>> remark that is even close to meaning what you say.

False.

Not an argument one can counter. Well done!

I'll take a crack at it, if I may.

True.

That is: In the original paper, he never uses the words "input" or "decidable".

(He doesn't even use the word 'halt'.)

Indeed, but you are assuming that someone called Mr Flibble is posting according to the usual conventions. I would not want to assume that his "false" relates to the immediately preceding quoted text. I fact, I
would not want to assume that anything he writes relates to anything at
all in the post being replied to. I think he likes posting
technical-sounding words. He certainly has no intention to communicate
using words with agreed-upon meanings.

--
Ben.

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On 5/16/25 5:47 PM, olcott wrote:

On 5/16/2025 4:44 PM, olcott wrote:

On 5/16/2025 4:33 PM, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Fri, 16 May 2025 00:59:02 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh?  Partial deciders are a thing.

Yes.  That does not alter the fact that no input is undecidable.

Pathological input is undecidable as pathological input is an
"impossible
program" [Strachey 1965].

The most likely explanation is that you don't know what decidable means. >>> Either that or you just like posting remarks for the sake of it.

Sure and these two PhD computer science professors
would also have no idea what the

TERMS OF THEIR ART MEAN.

Right, they clearly don't.

Their writing show they don't understand the meaning of a Program in the
field.

PHD sometimes mean just they Piled it Higher and Deeper.

Computer Science is a large field, and many people in it never studied
this corner of Computation Theory, as it actually has little application
to most programing exercises, as it intentionally ignores a lot of
asspects that are very important to application programming.

Problems with the Halting Problem
Eric C.R. Hehner
https://www.cs.toronto.edu/~hehner/PHP.pdf

Halting misconceived?
Bill Stoddart
August 25, 2017
https://www.complang.tuwien.ac.at/anton/euroforth/ef17/papers/
stoddart.pdf

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On 5/16/25 5:44 PM, olcott wrote:

On 5/16/2025 4:33 PM, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Fri, 16 May 2025 00:59:02 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

On Thu, 15 May 2025 13:23:43 +0100, Ben Bacarisse wrote:

Mr Flibble <flibble@red-dwarf.jmc.corp> writes:

the truth is pathlogical input is undecidable:

No input[1] is undecidable.

Eh?  Partial deciders are a thing.

Yes.  That does not alter the fact that no input is undecidable.

Pathological input is undecidable as pathological input is an
"impossible
program" [Strachey 1965].

The most likely explanation is that you don't know what decidable means.
Either that or you just like posting remarks for the sake of it.

Sure and these two PhD computer science professors
would also have no idea what the terms of their are mean:

Problems with the Halting Problem
Eric C.R. Hehner
https://www.cs.toronto.edu/~hehner/PHP.pdf

Halting misconceived?
Bill Stoddart
August 25, 2017 https://www.complang.tuwien.ac.at/anton/euroforth/ef17/papers/stoddart.pdf

Yes, they show that they have no idea about computation theory, as they
make rookie mistakes, like not understand that programs will do what
they do. (They may be good in other parts of Computer Science, just not Computation Theory)

That just shows the problem with trying to base an argument on
"authority". It may work in general Philosophy, but not in a formal
system like Computation Theory.

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On 2025-05-16 15:34:50 +0000, olcott said:

On 5/16/2025 1:54 AM, Mikko wrote:

On 2025-05-15 15:33:01 +0000, Mr Flibble said:

On Thu, 15 May 2025 10:13:50 -0500, olcott wrote:

On 5/15/2025 1:27 AM, Mr Flibble wrote:

Peter is right to say that the halting problem as defined is flawed: he >>>>> agrees with me that there is category error at the heart of the problem >>>>> definition whereby the decider is conflated with the program being
analysed in an ill-formed self-referential dependency that manifests in >>>>> his simulating halt decider as "aborted" infinite recursion.

Peter however is wrong to say that aborting his infinite recursion is >>>>> equivalant to a halting state of non-halting: the truth is pathlogical >>>>> input is undecidable: that part Turing et al got right.

/Flibble

Introduction to the Theory of Computation 3rd Edition by Michael Sipser >>>> (Author)
4.4 out of 5 stars    568 ratings

https://www.amazon.com/Introduction-Theory-Computation-Michael- Sipser/dp/ >>> 113318779X


<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
If simulating halt decider H correctly simulates its input D until
H correctly determines that its simulated D would never stop
running unless aborted then

H can abort its simulation of D and correctly report that D
specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022> >>>>
HHH does correctly reject DDD and DD according to the exact meaning of >>>> the above words. It also seems to me that those words are a truism.

Sipser is wrong: he is disagreeing with Turing et al that pathological
input is undecidable.

Which sentence of Sipser contradicts which sentence of Turing?
Why do you think that Sipser is wrong and not Turing?

A simulating halt decider (SHD) (as defined above)
proves that there cannot be any input that actually
does the opposite of whatever value that its SHD
returns.

int main()
{
DDD(); // The HHH called by DDD() cannot report on its caller.
}

It does not matter what it proves unless you either post the proof
or post a proof that it proves what you claim it proves.

--
Mikko

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