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Simulating Halt Deciders

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olcott

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Aug 12, 2022, 11:45:43 AM8/12/22
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A simulating halt decider simulates its input until it correctly proves
that this simulation would never stop unless aborted.

typedef void (*ptr)();
int H(ptr p, ptr i); // simulating halt decider

void P(ptr x)
{
int Halt_Status = H(x, x);
if (Halt_Status)
HERE: goto HERE;
return;
}

int main()
{
Output("Input_Halts = ", H(P, P));
}

(a) I believe that most computer scientists would agree that a halt
decider must compute the mapping from its inputs to an accept or reject
state on the basis of the actual behavior specified by this input.

(b) Furthermore ALL computer scientists would agree that the correct
simulation of a machine description does derive the correct behavior
specified by this machine description.

This logically entails that when H(P,P) simulates its input until the
behavior of this input essentially matches the infinite recursion
behavior pattern, then H has correctly determined the halt status of P.

(a) The invoked H(P,P) simulates P(P) that calls a simulated H(P,P)
(b) that simulates P(P) that calls a simulated H(P,P)
(c) that simulates P(P) that calls a simulated H(P,P)...
Until H aborts the simulation of its input

*Halting problem proofs refuted on the basis of software engineering* ?
https://www.researchgate.net/publication/361701808_Halting_problem_proofs_refuted_on_the_basis_of_software_engineering


--
Copyright 2022 Pete Olcott

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

Mr Flibble

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Aug 12, 2022, 12:22:29 PM8/12/22
to
Nope. I have shown that a simulating halt decider needn't be recursive
in nature and you are incorrect to map your erroneous infinite recursion
to non-halting if P actually halts when run.

/Flibble

olcott

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Aug 12, 2022, 1:07:09 PM8/12/22
to
A simulating halt decider is never recursive in nature unless its input
is recursive:

void Infinite_Recursion(int N)
{
Infinite_Recursion(N);
}

int main()
{
Output("Input_Halts = ", H((u32)Infinite_Recursion, 0x777));

Mr Flibble

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Aug 12, 2022, 1:14:58 PM8/12/22
to
P calling H does not mean that P is recursive in nature, it means your
decider is recursive in nature.

/Flibble


olcott

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Aug 12, 2022, 1:32:47 PM8/12/22
to
A simulating halt decider always only simulates the behavior that its
input specifies.

Mr Flibble

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Aug 12, 2022, 1:46:25 PM8/12/22
to
On Fri, 12 Aug 2022 12:32:41 -0500
Again: H calling P calling H calling P does NOT mean that P is
recursive, it means H is recursive.

/Flibble

olcott

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Aug 12, 2022, 2:07:36 PM8/12/22
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Because H merely simulates whatever behavior that its input specifies
then it is P and *NOT* H that is recursive.

Mr Flibble

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Aug 12, 2022, 2:12:56 PM8/12/22
to
On Fri, 12 Aug 2022 13:07:30 -0500
WRONG. H is recursive not P, P is oblivious to how H decides P's
halting status, it is an implementation detail of *your* decider to
recursively call P and my SHD has no such recursion to "abort".

/Flibble

olcott

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Aug 12, 2022, 3:10:43 PM8/12/22
to
Because P calls H(P,P) that makes the simulated P the source of the
recursion.

> it is an implementation detail of *your* decider to

Always correctly simulate the behavior specified by its inputs.

> recursively call P and my SHD has no such recursion to "abort".
>
> /Flibble
>


Mr Flibble

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Aug 12, 2022, 3:21:42 PM8/12/22
to
Nope, it is an implementation detail of H to recursively call P in the
following sequence:

H calls P calls H calls P

Not all deciders are SHDs and it is for that reason that it is
incorrect to say that P is the source of the recursion: your H is the
source of the recursion.

>
> > it is an implementation detail of *your* decider to
>
> Always correctly simulate the behavior specified by its inputs.

Which your H doesn't do if P is defined as:

void P(ptr x)
{
int Halt_Status = H(x, x);
(void)Halt_Status;
return;
}

This P halts even though it references H.

/Flibble

olcott

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Aug 12, 2022, 3:29:41 PM8/12/22
to
When H(P,P) correctly simulates its first argument it never stops
running unless and until H aborts its simulation.

The ALWAYS means that the first argument specifies a non-halting
sequence of instructions.

Mr Flibble

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Aug 12, 2022, 5:33:09 PM8/12/22
to
On Fri, 12 Aug 2022 14:29:34 -0500
Except it isn't correctly simulating its first argument so cannot reach
a correct decision of halting or non-halting.

>
> The ALWAYS means that the first argument specifies a non-halting
> sequence of instructions.

Nope. Only your broken H suffers from infinite recursion which you
erroneously map to non-halting.

/Flibble

Richard Damon

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Aug 12, 2022, 5:52:17 PM8/12/22
to
On 8/12/22 11:45 AM, olcott wrote:
> A simulating halt decider simulates its input until it correctly proves
> that this simulation would never stop unless aborted.

And you H fails to meet that definiton as it stop its simulation and
returns non-halting to the call of H(P,P) before it has actually proven
that answer.

>
> typedef void (*ptr)();
> int H(ptr p, ptr i); // simulating halt decider
>
> void P(ptr x)
> {
>   int Halt_Status = H(x, x);
>   if (Halt_Status)
>     HERE: goto HERE;
>   return;
> }
>
> int main()
> {
>   Output("Input_Halts = ", H(P, P));
> }
>
> (a) I believe that most computer scientists would agree that a halt
> decider must compute the mapping from its inputs to an accept or reject
> state on the basis of the actual behavior specified by this input.

And the behavior of the input to H(P,P) is the behavior of P(P) BY
DEFINITION.

>
> (b) Furthermore ALL computer scientists would agree that the correct
> simulation of a machine description does derive the correct behavior
> specified by this machine description.

Right, the correct and COMPLET simulation of the input, using ALL the
behavior thus specifed.

>
> This logically entails that when H(P,P) simulates its input until the
> behavior of this input essentially matches the infinite recursion
> behavior pattern, then H has correctly determined the halt status of P.

No, because your "essentially" doesn't mean Correctly.

>
> (a) The invoked H(P,P) simulates P(P) that calls a simulated H(P,P)
> (b) that simulates P(P) that calls a simulated H(P,P)
> (c) that simulates P(P) that calls a simulated H(P,P)...
> Until H aborts the simulation of its input

But that isn't the actual behavior that your defined H would do.

So you are caught in your lie.

Pancho Valvejob

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Aug 13, 2022, 5:07:44 PM8/13/22
to
On 8/12/2022 8:45 AM, olcott wrote:
> A simulating halt decider simulates its input until it correctly proves

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Thanks!

Mikko

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Aug 14, 2022, 6:08:05 AM8/14/22
to
On 2022-08-12 15:45:34 +0000, olcott said:

> A simulating halt decider simulates its input until it correctly proves
> that this simulation would never stop unless aborted.

Can you prove that a simulating halt decider does something useful?
Can you prove that a simulating halt decider does something interesting?
Can you prove that a simulating halt decider is a halt decider?
Can you prove that a simulating halt decider is a decider?
Can you prove that a simulating halt decider exists?

Mikko

olcott

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Aug 14, 2022, 9:43:27 AM8/14/22
to
See for yourself.

*I am working on getting this to compile under Linux Ubuntu 16.04*

https://www.liarparadox.org/2022_07_22.zip
This is the complete system that compiles under:

Microsoft Visual Studio Community 2017
https://visualstudio.microsoft.com/vs/older-downloads/

Mikko

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Aug 15, 2022, 8:56:51 AM8/15/22
to
On 2022-08-14 13:43:21 +0000, olcott said:

> On 8/14/2022 5:08 AM, Mikko wrote:
>> On 2022-08-12 15:45:34 +0000, olcott said:
>>
>>> A simulating halt decider simulates its input until it correctly proves
>>> that this simulation would never stop unless aborted.
>>
>> Can you prove that a simulating halt decider does something useful?
>> Can you prove that a simulating halt decider does something interesting?
>> Can you prove that a simulating halt decider is a halt decider?
>> Can you prove that a simulating halt decider is a decider?
>> Can you prove that a simulating halt decider exists?
>>
>> Mikko
>>
>
> See for yourself.

I see. You can't.

Mikko

olcott

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Aug 15, 2022, 10:47:06 AM8/15/22
to
I am sick of people contradicting the verified facts.
I am not going to play that head game any more.
This code proves all those points.

https://www.liarparadox.org/2022_07_22.zip
This is the complete system that compiles under:

Microsoft Visual Studio Community 2017
https://visualstudio.microsoft.com/vs/older-downloads/

Here is the corresponding paper:
*Halting problem proofs refuted on the basis of software engineering* ?
https://www.researchgate.net/publication/361701808_Halting_problem_proofs_refuted_on_the_basis_of_software_engineering



Richard Damon

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Aug 15, 2022, 6:59:24 PM8/15/22
to
And that H is incorrect as H(P,P) will return 0, saying its input, which
is the specification of P(P) will never halt, while P(P) Does Halt.

This is clear by your construction of P and its definition.

Non-halting is not Halting, so H is wrong.


PERIOD.

You are just shown to be a liar or an idiot,

Take your choice,

It is proven, by the plain meaning of the words, which you don't seem to
understand or are intentionally not looking at.

Mikko

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Aug 18, 2022, 4:09:03 AM8/18/22
to
On 2022-08-15 14:46:59 +0000, olcott said:

> I am sick of people contradicting the verified facts.

Such as:
If H(P,P) = 0 and P(P) halts then H is not a halt decider.

Mikko

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