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From: olcott <polcott333@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: DD simulated by HHH cannot possibly halt (Halting Problem)
Date: Sat, 5 Apr 2025 14:01:58 -0500
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On 4/5/2025 1:45 PM, Richard Heathfield wrote:
> On 05/04/2025 19:11, olcott wrote:
>> On 4/5/2025 11:25 AM, dbush wrote:

>>> On 4/5/2025 11:59 AM, olcott wrote:>>>>
>>>> Introduction to the Theory of Computation 3rd Edition
>>>> by Michael Sipser (Author) (best selling textbook)
>>>>
>>>> <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 not what you think he agreed to:
>>>
>>>

void DDD()
{
   HHH(DDD);
   return;
}

>>
>> You have to show that by showing the details of how
>> what he agreed to is not accurately paraphrased by
>> *Simulating termination analyzer Principle*
> 
> No, you have to show firstly that your H determines anything at all 
> about D's behaviour. 

First of all it is the concrete DDD and the hypothetical HHH.
As any C programmer can see DDD simulated by HHH would cause
any correct simulator to get stuck in recursive simulation.

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

The same applies to DD simulated by this same HHH.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer