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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Any honest person that knows the x86 language can see... predict
 correctly
Date: Wed, 31 Jul 2024 23:24:26 -0500
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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.
> 

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer