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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: Sufficient knowledge of C proves that DD specifies
 non-terminating behavior to HHH
Date: Sun, 9 Feb 2025 00:10:57 -0600
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On 2/8/2025 3:54 PM, Fred. Zwarts wrote:
> Op 08.feb.2025 om 15:47 schreef olcott:
>> On 2/8/2025 3:57 AM, Fred. Zwarts wrote:
>>> Op 08.feb.2025 om 06:53 schreef olcott:
>>>> On 2/7/2025 7:27 PM, Richard Damon wrote:
>>>>> On 2/7/25 8:12 PM, olcott wrote:
>>>>>> On 2/7/2025 5:56 PM, Richard Damon wrote:
>>>>>>> On 2/7/25 11:26 AM, olcott wrote:
>>>>>>>> On 2/7/2025 6:20 AM, Richard Damon wrote:
>>>>>>>>> On 2/6/25 10:02 PM, olcott wrote:
>>>>>>>>>> On 2/6/2025 8:21 PM, Richard Damon wrote:
>>>>>>>>>>> On 2/6/25 5:18 PM, olcott wrote:
>>>>>>>>>>>> On 2/6/2025 1:51 PM, Richard Damon wrote:
>>>>>>>>>>>>> On 2/6/25 1:26 PM, olcott wrote:
>>>>>>>>>>>>>> On 2/6/2025 10:52 AM, Bonita Montero wrote:
>>>>>>>>>>>>>>> Am 05.02.2025 um 16:11 schrieb olcott:
>>>>>>>>>>>>>>>> On 2/5/2025 1:44 AM, Bonita Montero wrote:
>>>>>>>>>>>>>>>>> Am 05.02.2025 um 04:38 schrieb olcott:
>>>>>>>>>>>>>>>>>> This treatment does not typically last very long and
>>>>>>>>>>>>>>>>>> will be immediately followed by a riskier fourth line
>>>>>>>>>>>>>>>>>> of treatment that has an initial success rate much higher
>>>>>>>>>>>>>>>>>> than its non progression mortality rate.
>>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>> Halting problem solved !
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> The halting problem proof input does specify non-halting
>>>>>>>>>>>>>>>> behavior to its decider.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> https://www.researchgate.net/ 
>>>>>>>>>>>>>>>> publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> LOOOOOOOOL
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Anyone that understands the C programming language
>>>>>>>>>>>>>> sufficiently well (thus not confused by the unreachable
>>>>>>>>>>>>>> "if" statement) correctly understands that DD simulated
>>>>>>>>>>>>>> by HHH cannot possibly reach its own return instruction.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>
>>>>>>>>>>>>> And anyone that understand the halting problem knows that 
>>>>>>>>>>>>> isn't the question being asked. The quesiton you NEED to 
>>>>>>>>>>>>> ask is will the program described by the input halt when run?
>>>>>>>>>>>>>
>>>>>>>>>>>>> Since you start off with the wrong question, you logic is 
>>>>>>>>>>>>> just faulty.
>>>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> Everyone that thinks my question is incorrect is wrong.
>>>>>>>>>>>> It has always been a mathematical mapping from finite
>>>>>>>>>>>> strings to behaviors. That people do not comprehend this
>>>>>>>>>>>> shows the shallowness of the depth of the learned-by-rote
>>>>>>>>>>>> (lack of) understanding.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> No, you are just incorreect as you don't know what you are 
>>>>>>>>>>> talking about.
>>>>>>>>>>>
>>>>>>>>>>> Yes, it is a mapping of the string to the behavior, and that 
>>>>>>>>>>> mapping is DEFINED to be the halting behavior of the program 
>>>>>>>>>>> the string describes.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> No this is incorrect. The input finite string specifies
>>>>>>>>>> (not merely describes) non halting behavior to its decider.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> No, since the definition of "Halting Behavior" is the behavior 
>>>>>>>>> of the progran being run.
>>>>>>>>>
>>>>>>>>
>>>>>>>> It may seem that way to people that have learned-by-rote
>>>>>>>> as their only basis. It is actually nothing like that.
>>>>>>>
>>>>>>> No, that *IS* the definition.
>>>>>>>
>>>>>>
>>>>>> A termination analyzer computes the mapping from finite
>>>>>> strings to the actual behavior that these finite strings
>>>>>> specify. That this is not dead obvious to everyone here
>>>>>> merely proves that learned-by-rote does not involve any
>>>>>> actual comprehension.
>>>>>>
>>>>>>
>>>>>
>>>>> And the behavior the finite string specifies is the behavior of 
>>>>> running the program. 
>>>>
>>>> That is verifiably factually incorrect.
>>>> The running program has a different execution trace
>>>> than the behavior that DD specifies to HHH.
>>>>
>>>
>>> If so, then it proves the failure of the simulation. The simulation 
>>> aborts too soon on unsound grounds, one cycle before the normal 
>>> termination of the program.
>>>
>>
>> This proves that you simply don't have sufficient
>> understanding of the C programming language.
>> DD simulated by HHH cannot possibly terminate normally
>> is a verified fact.
>>
> 
> Which proves that HHH fails to make a correct decision about DD's 
> halting behaviour. All other methods (direct execution, simulation by a 
> world class simulator, etc.) show that DD halts. But HHH fails to see 
> it. Everyone with sufficient understanding of programming sees that HHH 
> is not correctly programmed when it aborts one cycle before the 
> simulation would end normally.

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

int DD()
{
   int Halt_Status = HHH(DD);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}

int main()
{
   HHH(DD);
}

You lack the ability to do the execution trace
of HHH simulating DD calling HHH(DD) simulating DD...

If you have no idea what recursion is you will not be
able to understand what I am saying.

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