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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 10:22:45 +0200
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Op 10.mei.2025 om 05:24 schreef olcott:
> 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.
And the input specifies a program that aborts and halts, but HHH ignores 
that, so its mapping is incorrect.
It seems to difficult for you to accept verifiable facts hen they 
disturb your dreams.