Re: Any honest person that knows the x86 language can see... predict correctly is to say Halting,

On 8/1/24 8:30 AM, olcott wrote:

On 8/1/2024 6:34 AM, Richard Damon wrote:

On 7/31/24 11:51 PM, olcott wrote:

On 7/31/2024 10:08 PM, wij wrote:

On Tue, 2024-07-30 at 18:50 -0500, olcott wrote:

>
It is not supposed to be a general solution to the halting problem.
it only shows how the "impossible" input is correctly determined
to be non halting.
>

>
But how do you determine it is non-halting?
>
As I know you are even unable to define what 'halt' mean !!!
>

I have done this thousands of times and after someone
has read these thousands of times they say that I never
said it once.
>
void DDD()
{
   HHH(DDD);
   return;
}
>
int main()
{
   HHH(DDD);
}
>
If DDD correctly emulated by HHH cannot possibly
reach its return instruction then it never halts.
>
>

>
But only *IF*  HHH *DOES* correctly emulate its input, which means it can't abort its emulation,

 *No stupid it has never meant that*
 <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     *simulating halt decider H correctly simulates its input D*
     *until H correctly determines that its simulated D would never*
     *stop running unless aborted*
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
 

Really, do you have a source for that?
I say it does, and that follows from the DEFINITION of the only thing described in Computation theory as a "correct simulator", which is the UTM, which BY DEFINITION behaves EXACTLY like the machine it is simulating, even if that machine is non-halting, thus it can not abort its simulation.
You r claim is just another your your uncounted false claims that you have absolutely no support for, and don't even try to justify, because you know you are just making them up.
Sorry, you are just proving yourself to be a LIAR.
I will say, your subject at least is honest this time, the CORRECT answer for HHH is to say Halting, unfortunatly, that isn't what it says.