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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,comp.lang.c,comp.lang.c++
Subject: Re: HHH(DDD)==0 is correct
Followup-To: comp.theory
Date: Thu, 3 Jul 2025 10:17:53 -0500
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On 7/3/2025 9:50 AM, Richard Damon wrote:
> On 7/3/25 10:39 AM, olcott wrote:
>> On 7/3/2025 9:16 AM, Richard Damon wrote:
>>> On 7/2/25 10:50 PM, olcott wrote:
>>>> On 7/1/2025 11:37 AM, Mr Flibble wrote:
>>>>> On Mon, 30 Jun 2025 21:12:48 -0400, Richard Damon wrote:
>>>>>
>>>>>> On 6/30/25 2:30 PM, Mr Flibble wrote:
>>>>>>
>>>>>> PO just works off the lie that a correct simulation of the input is
>>>>>> different than the direct execution, even though he can't show the
>>>>>> instruction actually correctly simulated where they differ, and thus
>>>>>> proves he is lying.
>>>>>>
>>>>>> The closest he comes is claiming that the simulation of the "Call 
>>>>>> HHH"
>>>>>> must be different when simulated then when executed, as for "some
>>>>>> reason" it must be just because otherwise HHH can't do the 
>>>>>> simulation.
>>>>>>
>>>>>> Sorry, not being able to do something doesn't mean you get to 
>>>>>> redefine
>>>>>> it,
>>>>>>
>>>>>> You ar4e just showing you are as stupid as he is.
>>>>>
>>>>> No. A simulator does not have to run a simulation to completion if 
>>>>> it can
>>>>> determine that the input, A PROGRAM, never halts.
>>>>>
>>>>> /Flibble
>>>>
>>>> The most direct way to analyze this is that
>>>> HHH(DDD)==0 and HHH1(DDD)==1 are both correct
>>>> because DDD calls HHH(DDD) in recursive simulation and
>>>> DDD does not call HHH1(DDD) in recursive simulation.
>>>
>>> Nope. It seems you don't understand what the question actually IS 
>>> because you have just lied to yourself so much that you lost the 
>>> understanding of the queiston.
>>>
>>>>
>>>> *I can't imagine how Mike does not get this*
>>>
>>> I can't understand
>>>
>>>>
>>>> *Context of above dialogue*
>>>> *Context of above dialogue*
>>>> *Context of above dialogue*
>>>
>>> Context of your context:
>>>
>>> A Halt Decider is supposed to decide if the program given to it (via 
>>> some correct representation) will halt when run.
>>>
>>> Thus, "the input" needs to represent a program
>>>
>>>>
>>>> typedef void (*ptr)();
>>>> int HHH(ptr P);
>>>>
>>>> void DDD()
>>>> {
>>>>    HHH(DDD);
>>>>    return;
>>>> }
>>>>
>>>> int main()
>>>> {
>>>>    HHH(DDD);
>>>> }
>>>
>>> Which, by itself, isn't a valid input, or program. as HHH is undefined.
>>>
>>> Each different definition of HHH, gives a different problem.
>>>
>>> Your "logic" seems to be based on trying to re-define what a program 
>>> is, which just makes it a lie.
>>>
>>> "Programs" must be complete and self-contained in the field of 
>>> computability theory, something you don't seem to understand.
>>>
>>>>
>>>> Termination Analyzer HHH simulates its input until
>>>> it detects a non-terminating behavior pattern. When
>>>> HHH detects such a pattern it aborts its simulation
>>>> and returns 0. (HHH1 has identical code)
>>>>
>>>>
>>>
>>> But it CAN'T simulate the above input. as it isn't valid.
>>>
>>> You need to add the code of HHH to the input to let HHH simulate "the 
>>> input" to get anything.
>>>
>>
>> No I do not. The above paragraph has every detail that is needed.
> 
> Then how do you correctly simulate something you do not have.
> 
> Note, your "description" of HHH is just incorrect, as it is also 
> incomplete.
> 
> Simulating a LIE just gives you a lie.
> 
>>
>>> And at that point, you have different inputs for different HHHs, and 
>>> possibly different behaviors, which you logic forgets to take into 
>>> account, which just breaks it.
>>>
>>
>> Wrong.
>> It is because the what I specified does take this
>> into account that HHH(DDD)==0 and HHH1(DDD)==1 are correct.
> 
> Nope, becausee it violates the DEFINITION of what it means to simulate 
> something.
> 

*You don't even know what you mean by this*
What I mean is the execution trace that is derived
within the semantics of the C programming language.

HHH1 simulates DDD that calls HHH(DDD) that
simulates DDD and then simulates itself simulating DDD.
Then HHH(DDD) aborts its simulation and returns to the
DDD simulated by HHH1. Then this DDD returns to HHH1.

*Claude (smarter than ChatGPT) figured this out on its own*
https://claude.ai/share/da9b8e3f-eb16-42ca-a9e8-913f4b88202c

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer