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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: What is the correct halt status for HHH(DDD) ? --- Tautology
Date: Sat, 13 Jul 2024 15:40:01 -0400
Organization: i2pn2 (i2pn.org)
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On 7/13/24 3:23 PM, olcott wrote:
> 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.
> 

And you ignore the fundamental fact that ALL DDDs built on the same code 
in HHH will behave the same.

If you disagree, what instruction in one of the behaves differently (or 
is "correctly emulated" differently) in one than another.

You just don't understand this fundamental fact, which you just lie and 
claim to be untrue, but have never actually show the diffference.

The closes is by INCORRECTLY simulating the CALL H instruction to do 
something that isn't actually what a call H instruction does, in other 
words, you just LIE.

This just proves that you insufficiently understand the basics of how 
cojputers work to make any of your claims.

Go ahead, try to show where the differ at an actually correctly 
simulated instruction, and not just were you ASSUME they must differ 
based on the assumption that they can differ and they need to give 
different results.