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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 19:20:55 -0400
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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