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

On 5/16/25 11:00 AM, 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?
 

Well, as soon as HHH is defined to not get itself stuck in the loop, and thus not be the correct simulator of the input, that the correct simulaiton of the input, that references that actual HHH will reach the final state.
So, how many times, FOREVER, as any less and it doesn't correct simulate the input.
The fact that this makes it not a decider just shows that simulating halt deciders can't be both a correct simulator and a correct halt decider for all inputs (or even this one input).
Sorry, your truth fairy can't help you with this, as you are stuck in reality here.