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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Sat, 10 May 2025 10:25:28 +0200
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Op 10.mei.2025 om 06:13 schreef olcott:
> On 5/9/2025 10:53 PM, Richard Damon wrote:
>> On 5/9/25 11:24 PM, 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.
>>>
>>
>> The problem is you don't compute the correct mapping, as you LIE by 
>> changing the defined mapping with your strawman,
>>
>> Termination Analyzer, BY THEIR DEFINITION, determine if the program 
>> represented by their input will halt when run. PERIOD.
>>
> 
> When definitions contradict each other at
> least one of them is wrong.
> 
> HHH(DDD) computes the mapping from its input finite
> string according to the rules of the x86 language
> thus obtains the actual behavior THAT THIS INPUT STRING SPECIFIES.
> 
But ignores the most important part of the input that specifies that the 
program aborts and halt. So, the mapping is incorrect. A correct mapping 
uses the full input, not only an irrelevant part of the finite string.
It seems that you are unable to accept verifiable facts when they 
disturb your dreams.