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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 22:24:43 -0500
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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.

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