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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Simulating termination analyzers for dummies
Date: Mon, 17 Jun 2024 21:04:40 -0500
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On 6/17/2024 8:24 PM, Richard Damon wrote:
> On 6/17/24 9:16 PM, olcott wrote:
>> On 6/17/2024 5:42 PM, Richard Damon wrote:
>>> On 6/17/24 8:20 AM, olcott wrote:
>>>> On 6/17/2024 3:31 AM, Fred. Zwarts wrote:
>>>>> Op 17.jun.2024 om 05:33 schreef olcott:
>>>>>> To understand this analysis requires a sufficient knowledge of
>>>>>> the C programming language and what an x86 emulator does.
>>>>>>
>>>>>> Unless every single detail is made 100% explicit false assumptions
>>>>>> always slip though the cracks. This is why it must be examined at
>>>>>> the C level before it is examined at the Turing Machine level.
>>>>>>
>>>>>> typedef void (*ptr)();
>>>>>> int H0(ptr P);
>>>>>>
>>>>>> void Infinite_Loop()
>>>>>> {
>>>>>>    HERE: goto HERE;
>>>>>> }
>>>>>>
>>>>>> void Infinite_Recursion()
>>>>>> {
>>>>>>    Infinite_Recursion();
>>>>>> }
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    H0(DDD);
>>>>>>    return;
>>>>>> }
>>>>>>
>>>>>> int main()
>>>>>> {
>>>>>>    H0(Infinite_Loop);
>>>>>>    H0(Infinite_Recursion);
>>>>>>    H0(DDD);
>>>>>> }
>>>>>>
>>>>>> Every C programmer that knows what an x86 emulator is knows that 
>>>>>> when H0
>>>>>> emulates the machine language of Infinite_Loop, 
>>>>>> Infinite_Recursion, and
>>>>>> DDD that it must abort these emulations so that itself can terminate
>>>>>> normally.
>>>>>>
>>>>>> When this is construed as non-halting criteria then simulating
>>>>>> termination analyzer H0 is correct to reject these inputs as non-
>>>>>> halting.
>>>>>>
>>>>>
>>>>> For Infinite_Loop and Infinite_Recursion that might be true, 
>>>>> because there the simulator processes the whole input.
>>>>>
>>>>> The H0 case is very different. For H0 there is indeed a false 
>>>>> assumption, as you mentioned. Here H0 needs to simulate itself, but 
>>>>> the simulation is never able to reach the final state of the 
>>>>> simulated self. The abort is always one cycle too early, so that 
>>>>> the simulating H0 misses the abort. Therefore this results in a 
>>>>> false negative.
>>>>> (Note that H0 should process its input, which includes the H0 that 
>>>>> aborts, not a non-input with an H that does not abort.)
>>>>>
>>>>> This results in a impossible dilemma for the programmer. It he 
>>>>> creates a H that does not abort, it will not terminate. 
>>>>
>>>> *Therefore what I said is correct*
>>>> When every input that must be aborted is construed as non-halting
>>>> then the input to H0(DDD) is correctly construed as non-halting.
>>>
>>> In other words, if you allow yourself to LIE, you can claim the wrong 
>>> answer is right.
>>>
>>> Since your "Needing to abort" is NOT the same as halting, all you are 
>>> doing is admitting that your whole logic system is based on the 
>>> principle that LIES ARE OK.
>>>
>>
>> "Needing to abort" <is> the same as a NOT halting input.
>> You are simply too ignorant to understand this.
>>
> 
> Nope, not if you are comparing DIFFERENT version of the input.
> 
It is ALWAYS the exact same sequence of bytes.

> 
>> When I explain this in terms of of mathematical mappings
>> from finite strings to behaviors this simply leaps over
>> everyone's head.
>>
> 
> As has been shown, you can apply the input to H0 (when you don't change 
> it, so the call to H0 still goes to this H0), to a UTM and it will reach 
> the final end, so *THIS* H0 did not "Need" to abort its input, but did 
> because it was programmend to.
> 

Not so much.

void DDD()
{
   UTM(DDD);
}

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer