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From: dbush <dbush.mobile@gmail.com>
Newsgroups: comp.lang.c
Subject: Re: Proving the: Simulating termination analyzer Principle
Date: Sat, 5 Apr 2025 21:22:23 -0400
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On 4/5/2025 8:43 PM, olcott wrote:
> On 4/5/2025 7:34 PM, dbush wrote:
>> On 4/5/2025 8:30 PM, olcott wrote:
>>> On 4/5/2025 5:27 PM, dbush wrote:
>>>> On 4/5/2025 6:18 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.
>>>>>
>>>>
>>>> Of course you're correct. His criteria is basically what happens if 
>>>> you replace the code of X with a pure simulator and run X(Y) for 
>>>> some Y.
>>>>
>>>
>>> Everyone else seems to think that the correct way
>>> to handle a pathological relationship between an
>>> input and a termination analyzer is to simply ignore
>>> the differences that this makes. THAT CAN'T BE RIGHT !!!
>>>
>>
>> Ignoring the relationship is exactly what you do when you change the 
>> code of HHH, thereby changing the input.
>>
>> Changing the input is not allowed.
> 
> Ignoring the fact that the pathological
> relationship between the input and the
> termination analyzer changes the behavior
> of the input is certainly incorrect.
> 
> Giving up and saying nothing can be done
> also seems incorrect.
> 
> What else is left?
> 

Realizing that the halting function is not a computable function, i.e no 
algorithm exists that computes the required mapping:


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