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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: Liar detector: Peter Olcott --- Ben's agreement
Date: Tue, 9 Jul 2024 22:51:55 -0400
Organization: i2pn2 (i2pn.org)
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On 7/9/24 10:14 AM, olcott wrote:
> On 7/9/2024 1:14 AM, Mikko wrote:
>> On 2024-07-08 17:36:58 +0000, olcott said:
>>
>>> On 7/8/2024 11:16 AM, Fred. Zwarts wrote:
>>>> Op 08.jul.2024 om 18:07 schreef olcott:
>>>>>
>>>>> Try to show how infinity is one cycle too soon.
>>>>>
>>>> You believe that two equals infinity.
>>>
>>> void Infinite_Loop()
>>> {
>>>    HERE: goto HERE;
>>> }
>>>
>>> void Infinite_Recursion()
>>> {
>>>    Infinite_Recursion();
>>> }
>>>
>>> void DDD()
>>> {
>>>    HHH(DDD);
>>> }
>>>
>>> Two cycles is enough to correctly determine that none
>>> of the above functions correctly emulated by HHH can
>>> possibly halt.
>>>
>>> That you don't see this is ignorance or deception.
>>
>> There is an important detail that determines whether an infinite
>> execution can be inferred. That is best illustrated by the following
>> examples:
>>
>> void Finite_Loop()
>> {
>>    int x = 10000;
>> HERE:
>>    if (x > 0) {
>>      x--;
>>      goto HERE;
>>    }
>> }
>>
>> void Finite_Recursion(int n)
>> {
>>    if (n > 0) {
>>      Finite_Recursion(n + 1);
>>    }
>> }
>>
>> void DDD()
>> {
>>    HHH(DDD); // HHH detects recursive simulation and then simulates no 
>> more
>> }
>>
>> The important difference is that in my examples there is a conditional
>> instruction that can (and does) prevent infinite exectuion.
>>
> 
> When we ask:
> Does the call from DDD emulated by HHH to HHH(DDD) return?
> We find that the emulated HHH never returns and this causes
> the emulated DDD to never halt.

No, the emulation of the emulated HHH never returns.

The emulated HHH does return, as all instances of a pure function with 
the same input behave the same.

You just don't understand the difference between reality and the 
observation of it.



> 
> _DDD()
> [00002163] 55         push ebp      ; housekeeping
> [00002164] 8bec       mov ebp,esp   ; housekeeping
> [00002166] 6863210000 push 00002163 ; push DDD
> [0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
> [00002170] 83c404     add esp,+04
> [00002173] 5d         pop ebp
> [00002174] c3         ret
> Size in bytes:(0018) [00002174]
> 
> DDD is correctly emulated by HHH which calls an emulated
> HHH(DDD) to repeat this process until the emulated DDD is
> aborted. At no point in this emulation does the call from
> DDD correctly emulated by HHH to HHH(DDD) ever return.
> 
> 
Yes, in no point of the EMULATION, that is the part of the behavior that 
the emuulator obeserves does that happen. It only happens in the part of 
the behavior that it doesn't observe, but which CAN be observed by using 
the direct exection or giving the FULL input to a real correct emulator 
that fully emulated the behavior of the FULL DDD (which includes that 
original HHH as part of it).