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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Every sufficiently competent C programmer knows --- Very Stupid
 Mistake or Liars
Date: Fri, 14 Mar 2025 17:14:01 -0500
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In-Reply-To: <ab3e264c48d27de208b12a2fdc6718c50ac61cf8@i2pn2.org>

On 3/14/2025 3:02 PM, joes wrote:
> Am Fri, 14 Mar 2025 10:58:39 -0500 schrieb olcott:
>> On 3/14/2025 10:55 AM, joes wrote:
>>> Am Fri, 14 Mar 2025 10:13:41 -0500 schrieb olcott:
>>>> On 3/14/2025 9:10 AM, Richard Damon wrote:
>>>>> On 3/13/25 11:53 PM, olcott wrote:
>>>>>> On 3/13/2025 10:03 PM, Richard Damon wrote:
>>>>>>> On 3/13/25 10:07 PM, olcott wrote:
>>>>>>>> On 3/13/2025 6:09 PM, Richard Damon wrote:
>>>>>>>>> On 3/13/25 9:41 AM, olcott wrote:
>>>>>>>>>> On 3/13/2025 6:18 AM, Dan Cross wrote:
>>>>>>>>>>> In article <vqud4e$36e14$3@dont-email.me>,
>>>>>>>>>>> Fred. Zwarts <F.Zwarts@HetNet.nl> wrote:
>>>>>>>>>>>> Op 12.mrt.2025 om 16:31 schreef olcott:
>>>
>>>>>>>> But a complete emulation can?
>>>>>>>>
>>>>>>> Yes, but an HHH that gives an answer doesn't do one, due to the
>>>>>>> pathological design of the template used to build DD to the HHH it
>>>>>>> calls (which is the only HHH that can exist, or you have violated
>>>>>>> the basic rules of programing and logic).
>>>>>>> We have two basic cases,
>>>>>>> 1) if HHH does the partial emulation you describe, then the
>>>>>>> complete emulation of DD will see that DD call HHH, and it will
>>>>>>> emulate its input for a while, then abort and theu return 0 to DD
>>>>>>> which will then halt.
>>>>>> int main()
>>>>>> {
>>>>>>      HHH(DDD); // No DDD can possibly ever return.
>>>>>> }
>>>>> Since HHH doesn;t call DDD, the statement is vacuous and shows a
>>>>> fundamental ignorance of what is being talked about.
> 
>>> Important distinction.
> 
> 
>>>>> Yes, No HHH can emulated DDD to the end, but since halting is DEFINED
>>>>> by the behavior of the program, and for every HHH that aborts and
>>>>> returns, the program of DDD, as tested with:
>>>>> int main()
>>>>> {
>>>>>       DDD()
>>>>> }
>>>>> will return to main, that shows that every HHH that returns 0 fails
>>>>> to be a Halt Decider or Termination Analyzer. PERIOD..
>>>> The only difference between HHH and HHH1 is that they are at different
>>>> locations in memory. DDD simulated by HHH1 has identical behavior to
>>>> DDD() executed in main().
> 
>>> Oh, I thought it was an unconditional simulator. The same code cannot
>>> produce different behaviour.
> 
> 
>>>> The semantics of the finite string input DDD to HHH specifies that it
>>>> will continue to call HHH(DDD) in recursive simulation.
>>>> The semantics of the finite string input DDD to HHH1 specifies to
>>>> simulate to DDD exactly once.
>>> No. DDD always specifies the one same thing.
>> counter-factual The semantics of the finite string input DDD to HHH
>> specifies that it will continue to call HHH(DDD) in recursive
>> simulation.
>> The semantics of the finite string input DDD to HHH1 specifies to
>> simulate to DDD exactly once.

> Uh no, there is only one call in either case. DD doesn't specify shit
> "to HHH"; it doesn't know about its runtime environment.

Even though DDD does not know it is calling its own
simulator and HHH does not know that DDD is calling
itself, it remains a verified fact that DDD simulated
by HHH does not halt because it is calling its own
simulator in recursive simulation.

> 
>> The only difference between HHH and HHH1 is that they are at different
>> locations in memory. DDD simulated by HHH1 has identical behavior to
>> DDD() directly executed in main().
> Now *that* is impossible if I swapped their addresses. Not that you
> could tell.
> 

If you swapped the address that is encoded in DDD
then HHH1(DDD) would return 0 and HHH(DDD) would
return 1.

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