On 10/20/2024 1:47 PM, Chris M. Thomasson wrote:
> On 10/19/2024 8:21 AM, Mike Terry wrote:
>> On 16/10/2024 21:41, Chris M. Thomasson wrote:
>>> On 10/15/2024 6:43 PM, Mike Terry wrote:
>>>> On 14/10/2024 23:17, Chris M. Thomasson wrote:
>>>>> On 10/13/2024 7:57 PM, olcott wrote:
>>>>>> On 10/13/2024 11:34 AM, Kaz Kylheku wrote:
>>>>>>> On 2024-10-12, Chris M. Thomasson
>>>>>>> wrote:
>>>>>>>> On 10/12/2024 11:28 AM, Janis Papanagnou wrote:
>>>>>>>>> On 12.10.2024 11:32, Jan van den Broek wrote:
>>>>>>>>>> 2024-10-12, Chris M. Thomasson
>>>>>>>>>> schrieb:
>>>>>>>>>>> On 10/11/2024 7:50 PM, olcott wrote:
>>>>>>>>>>
>>>>>>>>>> [Schnipp]
>>>>>>>>>>
>>>>>>>>>> As I see it, the main Halting Problem is Olcott not halting.
>>>>>>>>>
>>>>>>>>> LOL! - A very nice one. Thanks for that. :-)
>>>>>>>>
>>>>>>>> I second that. :^)
>>>>>>>
>>>>>>> You're likely thousand-seconding that. The Olcott not halting joke
>>>>>>> is many years old now, and will likely come up again.
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> My cancer has gotten worse.
>>>>>>
>>>>>> *ChatGPT explains why rebuttals of my work are incorrect*
>>>>>> https://chatgpt.com/share/6709e046-4794-8011-98b7-27066fb49f3e
>>>>>>
>>>>>> I had to dumb this down from the original halting problem
>>>>>> input so that reviewers can verify that HHH is correct
>>>>>> without hardly paying any attention at all:
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>> HHH(DDD);
>>>>>> return;
>>>>>> }
>>>>>>
>>>>>> When HHH is an x86 emulation based termination analyzer
>>>>>> then each DDD emulated by any HHH that it calls never returns.
>>>>> [...]
>>>>>
>>>>> Isn't that similar to:
>>>>>
>>>>> void foobar()
>>>>> {
>>>>> foobar();
>>>>> }
>>>>>
>>>>> ? >
>>>>
>>>> Similar, but different because HHH only performs a /partial/ step by
>>>> step emulation of DDD - it stops emulating after a while and
>>>> returns, so DDD() halts. foobar() will never halt (ignoring
>>>> physical resource constraints like running out of stack). foobar()
>>>> undergoes infinite recursive call. DDD() exhibits /finite/
>>>> recursive emulation, then halts.
>>>
>>> So, any similar to:
>>> ______________
>>> void foo(
>>> unsigned long ri,
>>> unsigned long rn
>>> ) {
>>> if (ri > rn) return;
>>> foo(ri + 1, rn);
>>> }
>>> ______________
>>>
>>> foo(0, 5);
>>>
>>> ?
>>
>> Yes - similar in that both :
>> - foo and PO's DDD exhibit a flavour of recursion, and
>> - both foo and DDD break the recursion at some point and subsequently
>> halt
>>
>> Different in that:
>> - foo exhibits "recursive call" while DDD exhibits "recursive
>> simulation".
>>
>> When a call is made, the caller cannot receive control until the
>> callee returns. So with call recursion, for the pattern to break, the
>> break has to occur at the innermost call - like with your foo.
>>
>> When a simulation is made, the simulator is still in control, probably
>> running a loop that simulates each individual instruction of the
>> simulated computation. So recursive simulation may break in the same
>> way as your foo, namely from the inner simulation then percolating
>> back through the nested simulations until the outer simulation exits.
>> Also it can break by one of the simulations simply deciding not to
>> continue its simulation any more and quiting its instruction
>> simulation loop. That's what PO's DDD does.
>>
>> Another example: Suppose GGG is coded to simulate 10000000
>> instrucions of GGG then exit. That is more similar to DDD than your
>> foo, because like DDD it exhibits recursive simulation, and like DDD
>> the recursion breaks from the outside. Everybody agrees that GGG
>> terminates - would you be tempted to say that it is correct to say GGG
>> never halts because it "exhibits infinitely recursive behaviour"?
>> [meaning that the combined GGG emulation trace shows that simulated
>> GGG simulates the exact same GGG again]. I doubt that! (The whole
>> idea is beyond stupid, given we agree GGG halts.)
>
> Thanks for the highly informative information Mike!
>
> I appreciate it. :^)
>
Mike understands my code far better than anyone besides me.
Yet he and everyone else disagrees with my work on the basis
of their answer to this question that I asked ChatGPT 4.0:
Could it be correct for HHH(DDD) to report on the behavior
of the directly executed DDD()?
https://chatgpt.com/share/67158ec6-3398-8011-98d1-41198baa29f2
The first less than one page of this ChatGPT session is
the entire basis that ChatGPT 4.0 uses to evaluate my work.
*This basis is all C and correct software engineering*
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
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