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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Thu, 5 Sep 2024 12:37:56 +0200
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Op 04.sep.2024 om 15:08 schreef olcott:
> On 9/4/2024 4:50 AM, Fred. Zwarts wrote:
>> Op 03.sep.2024 om 15:29 schreef olcott:
>>> On 9/3/2024 2:15 AM, Fred. Zwarts wrote:
>>>> Op 03.sep.2024 om 00:22 schreef olcott:
>>>>> On 9/2/2024 12:52 PM, Fred. Zwarts wrote:
>>>>>> Op 02.sep.2024 om 18:38 schreef olcott:
>>>>>>> A halt decider is a Turing machine that computes
>>>>>>> the mapping from its finite string input to the
>>>>>>> behavior that this finite string specifies.
>>>>>>>
>>>>>>> If the finite string machine string machine
>>>>>>> description specifies that it cannot possibly
>>>>>>> reach its own final halt state then this machine
>>>>>>> description specifies non-halting behavior.
>>>>>>>
>>>>>>> A halt decider never ever computes the mapping
>>>>>>> for the computation that itself is contained within.
>>>>>>>
>>>>>>> Unless there is a pathological relationship between
>>>>>>> the halt decider H and its input D the direct execution
>>>>>>> of this input D will always have identical behavior to
>>>>>>> D correctly simulated by simulating halt decider H.
>>>>>>>
>>>>>>> *Simulating Termination Analyzer H Not Fooled by Pathological 
>>>>>>> Input D*
>>>>>>> https://www.researchgate.net/ 
>>>>>>> publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>>>>>>>
>>>>>>> A correct emulation of DDD by HHH only requires that HHH
>>>>>>> emulate the instructions of DDD** including when DDD calls
>>>>>>> HHH in recursive emulation such that HHH emulates itself
>>>>>>> emulating DDD.
>>>>>>
>>>>>> Indeed, it should simulate *itself* and not a hypothetical other 
>>>>>> HHH with different behaviour.
>>>>>> If HHH includes code to see a 'special condition' and aborts and 
>>>>>> halts, then it should also simulate the HHH that includes this 
>>>>>> same code and 
>>>>>
>>>>>
>>>>> DDD has itself and the emulated HHH stuck in recursive emulation.
>>>>>
>>>>> void DDD()
>>>>> {
>>>>>    HHH(DDD);
>>>>>    return;
>>>>> }
>>>>
>>>> It is not DDD. It is HHH that has the problem when trying to 
>>>> simulate itself.
>>
>> Olcott removed the proof that I am right:
>>
>>         int main() {
>>           return HHH(main);
>>         }
>>
>> where HHH halts, but claims that it does not halt. No DDD needed to 
>> prove that HHH reports false negatives.
>>
>> Since he cannot prove that I am wrong, he thinks an ad hominem attack 
>> will help.
>>
>>>
>>> It does this correctly yet beyond your intellectual capacity.
>>>
>>
>> Then he shows again the 'trace' of an incorrect simulation.
>>
>>> _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]
>>>
>>> Instructions from machine address 00002172 through
>>> machine address 0000217a are emulated.
>>>
>>> What instruction of DDD do you believe comes next?
>>
>> Assuming a correct simulation:
>> The next instruction would be that at 000015d2 in HHH. 
> 
>  >> What instruction of DDD do you believe comes next?
> 
> What instruction
> of DDD
> of DDD
> of DDD
> of DDD
> of DDD
> do you believe comes next?
> 
> 

Olcott's language processor is too weak to see that I gave the answer 
already, but he cut it away. So, I repeat:
In a correct simulation the next instruction after the call is in HHH 
(not in DDD) and, when HHH returns, the next one in *DDD* is the 
construction at 0000217f, followed by the instructions at 00002182 and 
00002183 after which DDD halts.
Such correct simulations are shown by the unmodified world class 
simulator and even by HHH1, but HHH fails to complete this simulation, 
because it aborts too soon.
HHH cannot possibly simulate itself correctly up to the end.