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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Fri, 9 May 2025 23:44:48 -0500
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On 5/9/2025 11:32 PM, wij wrote:
> On Fri, 2025-05-09 at 23:18 -0500, olcott wrote:
>> On 5/9/2025 10:43 PM, wij wrote:
>>> On Fri, 2025-05-09 at 22:24 -0500, olcott wrote:
>>>> On 5/9/2025 10:13 PM, wij wrote:
>>>>> On Fri, 2025-05-09 at 19:40 -0700, Keith Thompson wrote:
>>>>>> olcott <polcott333@gmail.com> writes:
>>>>>>> On 5/9/2025 4:40 PM, Richard Heathfield wrote:
>>>>>>>> On 09/05/2025 21:15, olcott wrote:
>>>>>>>>> On 5/9/2025 3:07 PM, Richard Heathfield wrote:
>>>>>>>>>> On 09/05/2025 20:46, olcott wrote:
>>>>>>>>>>> We have not begun to get into any of those points.
>>>>>>>>>>> We are only asking can DDD correctly simulated
>>>>>>>>>>> by any HHH that can exist ever reach its own
>>>>>>>>>>> "return" instruction.
>>>>>>>>>>
>>>>>>>>>> DDD can't be correctly simulated by itself (which is effectively
>>>>>>>>>> what you're trying to do when you fire up the simulation from
>>>>>>>>>> inside DDD).
>>>>>>>>>
>>>>>>>>> How the Hell did you twist my words to say that?
>>>>>>>> I haven't touched your words. What I have done is to observe that
>>>>>>>> DDD's /only/ action is to call a simulator. Since DDD isn't itself a
>>>>>>>> simulator, there is nothing to simulate except a call to a
>>>>>>>> simulator.
>>>>>>>> It's recursion without a base case - a rookie error.
>>>>>>>> HHH cannot successfully complete its task, because it never regains
>>>>>>>> control after the first recursion. To return, it must abort the
>>>>>>>> simulation, which means the simulation fails.
>>>>>>>>
>>>>>>>>> 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”.
>>>>>>>> On what grounds can you persuade an extraordinarily sceptical
>>>>>>>> readership that HHH 'correctly simulated' DDD?
>>>>>>>
>>>>>>> Any competent C programmer can see that
>>>>>>> the call from DDD to HHH(DDD) (its own simulator)
>>>>>>> is equivalent to infinite recursion.
>>>>>>>
>>>>>>> On 5/8/2025 8:30 PM, Keith Thompson wrote:
>>>>>>>> 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;
>>>>>>>>         }
>>>>>>>>
>>>>>>>> Then the return statement (which is unnecessary anyway) will never be
>>>>>>>> reached.  In practice, the program will likely crash due to a stack
>>>>>>>> overflow, unless the compiler implements tail-call optimization, in
>>>>>>>> which case the program might just run forever -- which also means the
>>>>>>>> unnecessary return statement will never be reached.
>>>>>>
>>>>>> I had not intended to post again, but I feel the need to make
>>>>>> a clarification.
>>>>>>
>>>>>> I acknowledged that the return statement would never be reached
>>>>>> *given the assumption* that HHH correctly simulates DDD.  Given
>>>>>> that assumption, a call to DDD() should be equivalent to a call
>>>>>> to HHH(DDD).
>>>>>>
>>>>>> I did not address whether the assumption is valid.  I merely
>>>>>> temporarily accepted it for the sake of discussion, just as I would
>>>>>> accept that if I were ten feet tall I would bump my head against
>>>>>> the ceiling in my house.
>>>>>>
>>>>>> The discussion I had with olcott did not reach the point of
>>>>>> discussing *how* HHH could correctly simulate DDD, or whether it
>>>>>> would even be logically possible for it to do so.  I also did not
>>>>>> address any issues of partial simulation, where olcott claims that
>>>>>> HHH can "accurately simulate" only a few x86 instructions rather
>>>>>> than simulating its entire execution.  I did not participate in
>>>>>> any discussion that would require knowledge of x86 machine or
>>>>>> assembly code.  (I have no doubt that I could learn x86 machine
>>>>>> and assembly code reasonably well if motivated to do so, but I am
>>>>>> not so motivated.)
>>>>>>
>>>>>> What I acknowledged was barely more than "if HHH correctly simulates
>>>>>> DDD, then HHH correctly simulates DDD".  (My understanding from
>>>>>> posts by others, whom I presume to be sufficiently knowledgeable,
>>>>>> is that HHH logically cannot accurately simulate DDD.)  I would
>>>>>> prefer that olcott refrain from using my words to support any of
>>>>>> his arguments beyond the scope of what he and I directly discussed.
>>>>>
>>>>> Don't know why you people stick on the 'simulation' stuff so long.
>>>>> The HP simply asks for such an H (in function form. POOH does not
>>>>> resemble TM):
>>>>>     H(D)=1 if D() halt.
>>>>>     H(D)=0 if D() not halt.
>>>>
>>>> My invention of a simulating termination
>>>> analyzer shows exactly how to compute the
>>>> mapping that the input that HHH(DD) specifies
>>>> into a correct answer for the halting problem's
>>>> otherwise impossible input.
>>>>
>>>> All rebuttals are based on failing to compute
>>>> this mapping correctly.
>>>>
>>>
>>> What is the correct mapping?
>>>
>>
>> _DDD()
>> [00002172] 55         push ebp      ; housekeeping
>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>> [00002175] 6872210000 push 00002172 ; push DDD
>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>> [0000217f] 83c404     add esp,+04
>> [00002182] 5d         pop ebp
>> [00002183] c3         ret
>> Size in bytes:(0018) [00002183]
>>
>> Computing the mapping of DDD emulated by HHH
>> according the the rules of the x86 language
>> to its behavior by HHH actually emulating DDD.
> 
> The above says you have no idea what the mapping is.
> 
>>> If POOH are not talking about the mapping:
>>>
>>>     H(D)=1 if D() halt.
>>>     H(D)=0 if D() not halt.
>>>
>>
>> The way that simulating termination analyzers process
>> their input by showing all of the steps of how the mapping
>> must be computed refutes the above simplistic view.
> 
> No (real) problem with that. But the HP asks:
> 
> H(D)=1 if D() halt.
> H(D)=0 if D() not halt.
> 
> You still evade the question: Is POO H anything to do with the HP?
> 

I have recently proven that the above requirements are incorrect.
That people consistently ignore this proof with pure bluster
is not actually any rebuttal at all.

>>> POOH is likely nothing to do with HP
>>>
>>
>>
> 
> 


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