Re: Any honest person that knows the x86 language can see... predict correctly

On Wed, 2024-07-31 at 23:24 -0500, olcott wrote:

On 7/31/2024 11:03 PM, wij wrote:

On Wed, 2024-07-31 at 22:51 -0500, 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.
 
 

 
That's right, HHH(DDD) as shown should never halt.
 

 
<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>
 
 

But The Halting Problem asks HHH to return 1 or 0 (so to speak, because you
don't know the detail).
 
Since HHH does not return 1 or 0 to answer the question, it is not a decider.
You are dealing with POO Problem.
 

 

GUR says the halting problem is undecidable.
You can pull your god to support your claim (like DJT?) again, but GUR says it won't work.