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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: What is the correct halt status for HHH(DDD) ? --- Tautology
Date: Sat, 13 Jul 2024 14:23:29 -0500
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On 7/13/2024 2:10 PM, Richard Damon wrote:
> On 7/13/24 2:53 PM, olcott wrote:
>> On 7/13/2024 1:33 PM, Richard Damon wrote:
>>> On 7/13/24 2:15 PM, olcott wrote:
>>>> On 7/13/2024 12:25 PM, Richard Damon wrote:
>>>>> On 7/13/24 12:48 PM, olcott wrote:
>>>>>> What is the correct halt status for an input to
>>>>>> a simulating termination analyzer that calls its
>>>>>> own termination analyzer?
>>>>>>
>>>>>> typedef void (*ptr)();
>>>>>> int HHH(ptr P);
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    HHH(DDD);
>>>>>> }
>>>>>>
>>>>>> int main()
>>>>>> {
>>>>>>    HHH(DDD);
>>>>>> }
>>>>>>
>>>>>
>>>>> Halting.
>>>>>
>>>>> Since HHH defined to be a termination analyzer, by that definition 
>>>>> it must return to its caller.
>>>>>
>>>>> Since DDD has no inputs, its behavior isn't affected by any inputs, 
>>>>> and thus DDD will halt for ALL input conditions, so 
>>>>
>>>> You are stupidly saying that Infinite_Loop() will halt because
>>>> it has no inputs.
>>>>
>>>
>>> Where did I say no input means halting?
>>>
>>> I said that since DDD has no inputs, a Termination analyizer doesn't 
>>> need to look over all inputs, as there are no inputs to affect it.
>>>
>>> Maybe you forget that a Termination Analyzer is not the same thing as 
>>> a Halt Decider.
>>>
>>> A Halt Decider determines if the given program will halt for the 
>>> given input, and needs to be given both (if the program takes an input).
>>>
>>> A Termination Analyzer determines if a given program will halt for 
>>> every possible input, and thus is only given the program, and not the 
>>> input to test.
>>>
>>>
>>> Note, for Infinite_Loop below, it IS possible for a simulator to 
>>> detect a condition in a finite amount of time that an unbounded 
>>> emulation would never halt, so can answer correctly non-halting.
>>>
>>> The problem with trying to apply that to DDD is it isn't true, give 
>>> the DDD that uses the HHH that tries to think it has determined it to 
>>> never halt to a actual pure emulator (and thus DDD still calls that 
>>> HHH that thinks it has determined that DDD will not halt and aborts 
>>> its emulation and returns) then the pure emulatior will see that HHH 
>>> make that decisioin and return to DDD which will return.
>>>
>>> Thus HHH can never actually make that same conclusion. You logic is 
>>> incorrrect as you presume the input changes with the emulator, but it 
>>> actually doesn't, it changes with the emulator you have stated is the 
>>> one that gives the correct answer, which you can't change in the 
>>> middle of the problem.
>>>
>>>> void Infinite_Loop()
>>>> {
>>>>    HERE: goto HERE;
>>>> }
>>>>
>>>>> for HHH to be a correct termination analysizer it needs to return 
>>>>> the value to indicate Halting.
>>>>>
>>>> Yes
>>>
>>> And thus, it WILL return that value to the DDD that calls it, and 
>>> that DDD will halt.
>>>
>>>>
>>>>> Your version 
>>>>
>>>> I am asking What is the correct halt status for HHH(DDD)
>>>> that at least one element of every possible pure function
>>>> HHH can provide.
>>>>
>>>
>>> The only correct answer it is able to give is Halt.
>>>
>>
>> Yet any input that must be aborted to prevent the non
>> termination of HHH necessarily specifies non-halting
>> behavior or it would never need to be aborted.
> 
> But the DDD that this HHH is given doesn't need to be aborted.
> 
> That is proven by HHH1(DDD).
> 

That is not the same DDD instance and you know it.
That is the DDD after HHH has aborted its DDD.

My code shows HHH1 creating a process context and
HHH creating two more different contexts.

HHH1 depends on HHH aborting its own different DDD
for the DDD of HHH1 to halt.

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