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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: Simulating termination analyzers by dummies --- What does halting
 mean?
Date: Wed, 19 Jun 2024 10:07:21 -0500
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On 6/19/2024 9:11 AM, Fred. Zwarts wrote:
> Op 19.jun.2024 om 15:11 schreef olcott:
>> On 6/19/2024 3:18 AM, Fred. Zwarts wrote:
>>> Op 18.jun.2024 om 19:25 schreef olcott:
>>>> On 6/18/2024 12:06 PM, joes wrote:
>>>>
>>>> void DDD()
>>>> {
>>>>    H0(DDD);
>>>> }
>>>>
>>>> DDD correctly simulated by any H0 cannot possibly halt.
>>>>
>>>>> DDD halts iff H0 halts.
>>>>
>>>> Halting is a technical term-of-the-art that corresponds
>>>> to terminates normally. Because Turing machines are
>>>> abstract mathematical objects there has been no notion
>>>> of abnormal termination for a Turing machine.
>>>>
>>>> We can derive a notion of abnormal termination for Turing
>>>> machines from the standard terms-of-the-art.
>>>>
>>>> Some TM's loop and thus never stop running, this is classical
>>>> non-halting behavior. UTM's simulate Turing machine descriptions.
>>>> This is the same thing as an interpreter interpreting the
>>>> source-code of a program.
>>>>
>>>> A UTM can be adapted so that it only simulates a fixed number
>>>> of iterations of an input that loops. When this UTM stops
>>>> simulating this Turing machine description we cannot correctly
>>>> say that this looping input halted.
>>>>
>>>
>>> If the code specifies 5 iterations and the simulator simulates only 3 
>>> iterations, it is incorrect to conclude that the repetition show 
>>> non-halting behaviour.
>>
>> It is correct do say that the simulated input did not terminate
>> normally, thus defining the notion of abnormal termination
>> within Turing machines.
>>
>>> Similarly, when your H, H0, or other H simulates itself, its 
>>> simulation aborts one cycle too early and therefore the non-halting 
>>> conclusion is incorrect.
>>
>> I was confused bout this for three days four years ago and then I
>> got over it. No simulating termination analyzer can wait for an
>> inner instance of itself to abort the simulation or it waits forever.
>>
>> Whenever the outer directly executed simulating termination analyzer
>> waits for any fixed number of repeating states it remains permanently
>> ahead of the next inner instance by exactly one repeating state. If
>> this is going to be permanently over your head then we need to stop
>> talking.
> 
> No, I understand it perfectly, but it seems to be over your head. We 
> agree that H needs to stop to prevent infinite recursion, but it is over 
> your head to see that when it stops, it misses the part of itself where 
> its simulation also stops, one repeating state further. So, the 
> non-halting conclusion is wrong, because the abort is premature.

typedef void (*ptr)();
int H0(ptr P);

void Infinite_Loop()
{
   HERE: goto HERE;
}

void Infinite_Recursion()
{
   Infinite_Recursion();
}

void DDD()
{
   H0(DDD);
}

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.

Every C programmer has agreed thus you simply don't know these things
well enough.

> So, your reasoning must lead to the only possible conclusion that a 
> simulator is unable to simulate itself properly, which causes false 
> negatives if its return value must be interpreted as a non-halting 
> behaviour. Instead, a return value of 'false' indicates an 'unable to 
> simulate' state of the simulator, which is not equivalent to 'non-halting'.

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