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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Thu, 12 Sep 2024 11:18:35 +0200
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Op 12.sep.2024 om 00:35 schreef olcott:
> On 9/11/2024 11:23 AM, Fred. Zwarts wrote:
>> Op 11.sep.2024 om 13:49 schreef olcott:
>>> On 9/10/2024 6:50 AM, Fred. Zwarts wrote:
>>>> Op 09.sep.2024 om 20:19 schreef olcott:
>>>>> On 9/8/2024 9:53 AM, Mikko wrote:
>>>>>> On 2024-09-07 13:57:00 +0000, olcott said:
>>>>>>
>>>>>>> On 9/7/2024 3:29 AM, Mikko wrote:
>>>>>>>> On 2024-09-07 05:12:19 +0000, joes said:
>>>>>>>>
>>>>>>>>> Am Fri, 06 Sep 2024 06:42:48 -0500 schrieb olcott:
>>>>>>>>>> On 9/6/2024 6:19 AM, Mikko wrote:
>>>>>>>>>>> On 2024-09-05 13:24:20 +0000, olcott said:
>>>>>>>>>>>> On 9/5/2024 2:34 AM, Mikko wrote:
>>>>>>>>>>>>> On 2024-09-03 13:00:50 +0000, olcott said:
>>>>>>>>>>>>>> On 9/3/2024 5:25 AM, Mikko wrote:
>>>>>>>>>>>>>>> On 2024-09-02 16:38:03 +0000, olcott said:
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> A halt decider is a Turing machine that computes the 
>>>>>>>>>>>>>>>> mapping from
>>>>>>>>>>>>>>>> its finite string input to the behavior that this finite 
>>>>>>>>>>>>>>>> string
>>>>>>>>>>>>>>>> specifies.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> A halt decider needn't compute the full behaviour, only 
>>>>>>>>>>>>>>> whether
>>>>>>>>>>>>>>> that behaviour is finite or infinite.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>> New slave_stack at:1038c4 Begin Local Halt Decider Simulation
>>>>>>>>>
>>>>>>>>>>>>>> Local Halt Decider: Infinite Recursion Detected Simulation 
>>>>>>>>>>>>>> Stopped
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Hence  HHH(DDD)==0 is correct
>>>>>>>>>>>>>
>>>>>>>>>>>>> Nice to see that you don't disagree with what said.
>>>>>>>>>>>>> Unvortunately I can't agree with what you say.
>>>>>>>>>>>>> HHH terminates,
>>>>>>>>>>>>> os DDD obviously terminates, too. No valid
>>>>>>>>>>>>
>>>>>>>>>>>> DDD emulated by HHH never reaches it final halt state.
>>>>>>>>>>>
>>>>>>>>>>> If that iis true it means that HHH called by DDD does not 
>>>>>>>>>>> return and
>>>>>>>>>>> therefore is not a ceicder.
>>>>>>>>>> The directly executed HHH is a decider.
>>>>>>>>> What does simulating it change about that?
>>>>>>>>
>>>>>>>> If the simulation is incorrect it may change anything.
>>>>>>>>
>>>>>>> PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
>>>>>>> PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
>>>>>>> PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
>>>>>>> PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
>>>>>>> PATHOLOGICAL RELATIONSHIPS CHANGE BEHAVIOR
>>>>>>
>>>>>> However, a correct simultation faithfully imitates the original
>>>>>> behaviour.
>>>>>>
>>>>>
>>>>> _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]
>>>>>
>>>>> A correct emulation obeys the x86 machine code even
>>>>> if this machine code catches the machine on fire.
>>>>>
>>>>> It is impossible for an emulation of DDD by HHH to
>>>>> reach machine address 00002183 AND YOU KNOW IT!!!
>>>>>
>>>>
>>>> It seems olcott also knows that HHH fails to reach the machine 
>>>> address 00002183, because it stop the simulation too soon.
>>>
>>> No the issue is the you insist on remaining too stupid
>>> to understand unreachable code.
>>>
>>> void Infinite_Recursion()
>>> {
>>>    Infinite_Recursion();
>>>    OutString("Can't possibly get here!");
>>
>> Olcott keeps dreaming of infinite recursions, even when HHH aborts 
>> after two cycles. Two is not infinite.
>>
> 
> Yet in this same way Infinite_Recursion() itself
> it not infinite when HHH aborts it in two cycles.
Olcott keeps dreaming of infinite recursions, when in fact there are 
only two recursions.

Olcott is very good in twisting the meaning of words. When a simulation 
stops halfway a halting program, it does not make sense to present a 
non-halting program.
HHH is more like

void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
   printf ("Olcott thinks this is never printed.\n");
}

> 
> What makes Infinite_Recursion() non-halting even
> when it stops being emulated is that it cannot
> possibly reach past its own first instruction.

Further stopping the simulation halfway the halting program does not 
prove that the remainder of the code is not reachable.
The remainder of the code of HHH and DDD is reachable, as proven by the 
direct execution, by the unmodified world class  simulator and even by HHH1.
Olcott thinks that if the simulation of Finite_Recursion is stopped 
after two recursions it proves that the next recursion is unreachable 
and therefore Finite_Recursion does not halt.

When will he finally realise that, although the return from HHH to DDD 
is reachable, HHH does not reach it because it stops too soon?
The fact that HHH does not reach the end, proves that the decision to 
stop was incorrect, not that the remainder is unreachable.
HHH cannot possibly simulate itself correctly up to the end.
The algorithm to decides that there is an infinite recursion, has been 
proven to have false positives. Just as in:

        int main() {
          return HHH(main);
        }

where HHH halts but decides that it detected an infinite recursion.