Re: Overcoming the proof of undecidability of the Halting Problem by a simple example in C

On 5/16/2025 10:11 AM, Mr Flibble wrote:

On Fri, 16 May 2025 10:00:26 -0500, olcott wrote:
 

On 5/16/2025 2:01 AM, Mikko wrote:

On 2025-05-15 21:11:35 +0000, olcott said:
>

On 5/15/2025 3:59 PM, Mr Flibble wrote:

On Thu, 15 May 2025 15:47:16 -0500, olcott wrote:
>

I overcome the proof of undecidability of the Halting Problem in
that the code that "does the opposite of whatever value that HHH
returns" becomes unreachable to DD correctly simulated by HHH.
>
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
>
HHH simulates DD that calls HHH(DD) to simulate itself again over
and over until HHH sees this repeating pattern and aborts or both
HHH and DD crash due to OOM error.

>
It is not possible for HHH to simulate DD because we are already
inside DD when we call HHH:

>
A partial simulation is possible. But at some point HHH discontinues
the simulation and returns a guessed answer.
>
>

void DDD()
{
    HHH(DDD);
    return;
}
>
int main()
{
    HHH(DDD);
}
>
HHH simulates DDD the simulated DDD calls HHH(DDD)
>
HHH simulates DDD the simulated DDD calls HHH(DDD)
>
HHH simulates DDD the simulated DDD calls HHH(DDD)
>
HHH simulates DDD the simulated DDD calls HHH(DDD)
>
HHH simulates DDD the simulated DDD calls HHH(DDD)
>
How many more times before the fact that DDD correctly simulated by HHH
cannot possibly reach its "return" statement?

 How many more times before you realise that this recursion is due to the
category error I have identified and as such it is undecidable.
 /Flibble

It would be a category error for HHH to report
anything to an input DD that actually does the
opposite of whatever value that HHH returns.
No such DD can possibly exist.
int main()
{
   DD(); // HHH cannot report on its caller.
}
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer