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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Sufficient knowledge of C proves that DD specifies
 non-terminating behavior to HHH
Date: Sun, 9 Feb 2025 12:04:25 -0500
Organization: i2pn2 (i2pn.org)
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On 2/9/25 10:33 AM, olcott wrote:
> On 2/9/2025 7:08 AM, Richard Damon wrote:
>> On 2/9/25 1:10 AM, olcott wrote:
>>> 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.
>>>
>>
>> No, YOU lack the understanding of what a program is.
>>
>> Your first problem is that function "DD" isn't a "program" by itself, 
>> but only becomes one when you include as part of it the code for HHH. 
>> And thus, the specific HHH that exists at this exact point IS HHH, and 
>> it can not be changed.
>>
> 
> It is this same way for every halting problem instance.
> It is an easily verified fact that DD cannot possibly reach
> its own "if" statement when-so-ever HHH is a simulating
> termination analyzer.


Which doesn't exist as any program that is a correct simulator, will not 
be a decider, and no program that is a decider can be a correct simulator.

> 
> The only reason that the halting problem proof has never
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