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From: olcott <polcott333@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: Proving the: Simulating termination analyzer Principle
Date: Sat, 5 Apr 2025 20:49:14 -0500
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On 4/5/2025 8:19 PM, dbush wrote:
> On 4/5/2025 8:36 PM, olcott wrote:
>> On 4/5/2025 6:20 PM, dbush wrote:
>>> On 4/5/2025 7:18 PM, olcott wrote:
>>>> On 4/5/2025 5:18 PM, Richard Heathfield wrote:
>>>>> On 05/04/2025 22:31, dbush wrote:
>>>>>> On 4/5/2025 5:29 PM, olcott wrote:
>>>>>>> On 4/5/2025 4:15 PM, dbush wrote:
>>>>>>>> On 4/5/2025 4:52 PM, olcott wrote:
>>>>>>>>> *Simulating termination analyzer Principle*
>>>>>>>>> It is always correct for any simulating termination
>>>>>>>>> analyzer to stop simulating and reject any input that
>>>>>>>>> would otherwise prevent its own termination.
>>>>>>>>>
>>>>>>>>> void DDD()
>>>>>>>>> {
>>>>>>>>>     HHH(DDD);
>>>>>>>>>     return;
>>>>>>>>> }
>>>>>>>>
>>>>>>>> Except when doing so would change the input, as is the case with 
>>>>>>>> HHH and DDD.
>>>>>>>>
>>>>>>>> Changing the input is not allowed.
>>>>>>>
>>>>>>> You may disagree that the above definition
>>>>>>> of simulating termination analyzer is correct.
>>>>>>>
>>>>>>> It is self-evident that HHH must stop simulating
>>>>>>> DDD to prevent its own non-termination.
>>>>>>>
>>>>>>
>>>>>> Changing the input is not allowed.
>>>>>
>>>>> You're right, but it doesn't matter very much as long as 
>>>>> terminates() *always* gets the answer right for any arbitrary 
>>>>> program tape and any data tape. Mr Olcott's fails to do that.
>>>>>
>>>>
>>>> Termination analyzers are not required to be infallible.
>>>>
>>>
>>>
>>> But they must still generate the required mapping for the input they 
>>> claim to answer correctly:
>>>
>>>
>>> Given any algorithm (i.e. a fixed immutable sequence of instructions) 
>>> X described as <X> with input Y:
>>>
>>> A solution to the halting problem is an algorithm H that computes the 
>>> following mapping:
>>>
>>> (<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
>>> (<X>,Y) maps to 0 if and only if X(Y) does not halt when executed 
>>> directly
>>>
>>
>> Exactly the opposite, they are only allowed to report
>> on what they see.
>>
> 
> No, they report what they are programed to report, which is some then
> computable function.
> 

If its a Turing computable function then it must
transform an input finite string into an output
according to some algorithm (defined sequence of steps).


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