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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Fri, 9 May 2025 14:25:49 -0400
Organization: i2pn2 (i2pn.org)
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On 5/9/25 12:04 PM, olcott wrote:
> On 5/9/2025 4:14 AM, Fred. Zwarts wrote:
>> Op 09.mei.2025 om 04:13 schreef olcott:
>>> On 5/8/2025 8:30 PM, Keith Thompson wrote:
>>>> olcott <polcott333@gmail.com> writes:
>>>>> On 5/8/2025 6:49 PM, Keith Thompson wrote:
>>>>>> olcott <polcott333@gmail.com> writes:
>>>>>> [...]
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>     HHH(DDD);
>>>>>>>     return;
>>>>>>> }
>>>>>>>
>>>>>>> If you are a competent C programmer then you
>>>>>>> know that DDD correctly simulated by HHH cannot
>>>>>>> possibly each its own "return" instruction.
>>>>>> "cannot possibly each"?
>>>>>> I am a competent C programmer (and I don't believe you can make
>>>>>> the same claim).  I don't know what HHH is.  The name "HHH" tells
>>>>>> me nothing about what it's supposed to do.  Without knowing what
>>>>>> HHH is, I can't say much about your code (or is it pseudo-code?).
>>>>>>
>>>>>
>>>>> For the purpose of this discussion HHH is exactly
>>>>> what I said it is. It correctly simulates DDD.
>>>>
>>>> Does HHH correctly simulate DDD *and do nothing else*?
>>>>
>>>> Does HHH correctly simulate *every* function whose address is passed
>>>> to it?  Must the passed function be one that takes no arguments
>>>> and does not return a value?
>>>>
>>>> Can HHH just *call* the function whose address is passed to it?
>>>> If it's a correct simulation, there should be no difference between
>>>> calling the function and "correctly simulating" it.
>>>>
>>>> My knowledge of C tells me nothing about *how* HHH might simulate
>>>> DDD.
>>>>
>>>
>>> HHH can only simulate a function that take no arguments
>>> and has no return value. HHH also simulates the entire
>>> chain of functions that this function calls. These can
>>> take arguments or not and have return values or not.
>>>
>>> Thus HHH ends up simulating itself (and everything
>>> that HHH calls) simulating DDD in an infinite
>>> sequence of recursive emulation until OOM error.
>>>
>>>>> We need not know anything else about HHH to
>>>>> know that DDD correctly simulated by HHH cannot
>>>>> possibly REACH its own "return" instruction.
>>>>
>>>> Assuming that HHH(DDD) "correctly simulates" DDD, and assuming it
>>>> does nothing else, your code would be equivalent to this:
>>>>
>>>>      void DDD(void) {
>>>>          DDD();
>>>>          return;
>>>>      }
>>>>
>>>
>>> Exactly. None of these people on comp.theory could
>>> get that even after three years.
>>>
>> Only if you forget that your proposed HHH aborts and returns.
> 
> *This slight augmentation takes that into account*
> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> When 1 or more statements of DDD are correctly
> simulated by HHH then this correctly simulated
> DDD cannot possibly reach its own “return statement”.
> 
> 
> 

But it can't simulate past the call to HHH, as it doesn't have the code 
for that as part of its input.

And partial simulation not reaching a final state is not "Non-Halting".