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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD simulated by HHH cannot possibly halt (Halting Problem)
Date: Sun, 13 Apr 2025 15:44:25 -0400
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On 4/13/2025 3:34 PM, olcott wrote:
> On 4/10/2025 6:27 PM, dbush wrote:
>> On 4/10/2025 7:26 PM, olcott wrote:
>>> On 4/10/2025 3:55 AM, Fred. Zwarts wrote:
>>>> Op 10.apr.2025 om 03:47 schreef olcott:
>>>>> On 4/9/2025 3:56 PM, dbush wrote:
>>>>>> On 4/9/2025 4:35 PM, olcott wrote:
>>>>>>> On 4/9/2025 1:58 PM, Fred. Zwarts wrote:
>>>>>>>> Op 09.apr.2025 om 19:29 schreef olcott:
>>>>>>>>>
>>>>>>>>> On 4/8/2025 10:31 AM, Fred. Zwarts wrote:
>>>>>>>>>> Op 08.apr.2025 om 17:13 schreef olcott:
>>>>>>>>>>> On 4/8/2025 2:45 AM, Fred. Zwarts wrote:
>>>>>>>>>>>> Op 08.apr.2025 om 06:33 schreef olcott:
>>>>>>>>>>>>>
>>>>>>>>>>>>> typedef void (*ptr)();
>>>>>>>>>>>>> int HHH(ptr P);
>>>>>>>>>>>>>
>>>>>>>>>>>>> int DD()
>>>>>>>>>>>>> {
>>>>>>>>>>>>>    int Halt_Status = HHH(DD);
>>>>>>>>>>>>>    if (Halt_Status)
>>>>>>>>>>>>>      HERE: goto HERE;
>>>>>>>>>>>>>    return Halt_Status;
>>>>>>>>>>>>> }
>>>>>>>>>>>>>
>>>>>>>>>>>>> int main()
>>>>>>>>>>>>> {
>>>>>>>>>>>>>    HHH(DD);
>>>>>>>>>>>>> }
>>>>>>>>>>>>>
>>>>>>>>>>>>> *Simulating termination analyzer Principle*
>>>>>>>>>>>>> It is always correct for any simulating termination
>>>>>>>>>>>>> analyzer to stop simulating and reject any input that
>>>>>>>>>>>>> would otherwise prevent its own termination.
>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> In this case there is nothing to prevent, because the finite 
>>>>>>>>>>>> string specifies a program that halts. 
>>>>>>>>>>>
>>>>>>>>>>> int DD()
>>>>>>>>>>> {
>>>>>>>>>>>    int Halt_Status = HHH(DD);
>>>>>>>>>>>    if (Halt_Status)
>>>>>>>>>>>      HERE: goto HERE;
>>>>>>>>>>>    return Halt_Status;
>>>>>>>>>>> }
>>>>>>>>>>>
>>>>>>>>>>> This stuff is simply over-your-head.
>>>>>>>>>>> HHH(DD) meets the above: *Simulating termination analyzer 
>>>>>>>>>>> Principle*
>>>>>>>>>>> Anyone with sufficient competence with the C programming 
>>>>>>>>>>> language
>>>>>>>>>>> will understand this.
>>>>>>>>>>>
>>>>>>>>>> Everyone with a little bit of C knowledge understands that if 
>>>>>>>>>> HHH returns with a value 0, then DDD halts. 
>>>>>>>>>
>>>>>>>>> DDD CORRECTLY SIMULATED BY HHH
>>>>>>>>> NOT ANY OTHER DAMN DDD IN THE UNIVERSE NITWIT.
>>>>>>>>>
>>>>>>>> If HHH would correctly simulate DD (and the functions called by 
>>>>>>>> DD) then the simulated HHH would return to DD and DD would halt.
>>>>>>>
>>>>>>> Simply over your level of technical competence.
>>>>>>>
>>>>>>>> But HHH failed to complete the simulation of the halting program, 
>>>>>>>
>>>>>>> HHH is only required to report on the behavior of its
>>>>>>> own correct simulation (meaning the according to the
>>>>>>> semantics of the C programming language) and would be
>>>>>>> incorrect to report on any other behavior.
>>>>>>
>>>>>> Which means HHH has conflicting requirements,
>>>>>
>>>>> No, it just means that the ones that you have
>>>>> been saying are f-cked up and no-one noticed this before.
>>>>>
>>>>>  > because to perform a
>>>>>  > correct simulation of its input it cannot halt itself, and 
>>>>> therefore
>>>>>  > can't report that.
>>>>> In other words you simply "don't believe in" the variant
>>>>> form of mathematical induction that HHH uses.
>>>>>
>>>>> A proof by induction consists of two cases. The first, the base 
>>>>> case, proves the statement for 𝑛=0 without assuming any knowledge 
>>>>> of other cases. The second case, the induction step, proves that if 
>>>>> the statement holds for any given case 𝑛=k, then it must also hold 
>>>>> for the next case 𝑛=k+1. These two steps establish that the 
>>>>> statement holds for every natural number 𝑛. The base case does not 
>>>>> necessarily begin with 𝑛=0, but often with 𝑛=1, and possibly with 
>>>>> any fixed natural number 𝑛=𝒩, establishing the truth of the 
>>>>> statement for all natural numbers 𝑛 ≥ 𝒩.   https:// 
>>>>> en.wikipedia.org/wiki/ Mathematical_induction
>>>>>
>>>> So the proof by induction shows that for any n HHH fails to complete 
>>>> the simulation. So, it has been proven that no HHH exists that is 
>>>> able to simulate correctly. It always aborts before it sees that the 
>>>> simulated HHH aborts as well.
>>>
>>> Yet again over-your-head.
>>> Unless the first HHH aborts
>>
>> Changing the input is not allowed.
> 
> int DD()
> {
>    int Halt_Status = HHH(DD);
>    if (Halt_Status)
>      HERE: goto HERE;
>    return Halt_Status;
> }
> 
> int main()
> {
>    HHH(DD);
> }
> 
> *Simulating termination analyzer Principle*
> It is always correct for any simulating termination
> analyzer to stop simulating and reject any input that
> would otherwise prevent its own termination.
> 

Except when doing so changes the input, as you're doing.

Changing the input is not allowed.

Note that copy/pasting the above in the future will be taken as 
repeating a previously refuted point, i.e. less than an answer, and your 
on-the-record admission that whatever point you used it to respond to is 
correct.

> The alternative is the  moronically stupid idea
> that termination analyzers should allow themselves
> to get get stuck in simulating an input that never halts.
> 

Yet you believe that the moronically stupid idea of changing the input 
is acceptable.

Changing the input is not allowed.

> Only internet trolls that don't giver a rat's ass for
> truth would suggest this nutty alternative.
> 

Like yourself, who thinks that a simulation analyzer should report on 
anything besides the halting function:


Given any algorithm (i.e. a fixed immutable sequence of instructions) X 
described as <X> with input Y:

A solution to the halting problem is an algorithm H that computes the 
following mapping:

(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly