Sujet : Re: Proving the: Simulating termination analyzer Principle
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.lang.cDate : 06. Apr 2025, 02:49:14
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vssmir$3m0q1$6@dont-email.me>
References : 1 2 3 4 5 6 7 8 9
User-Agent : Mozilla Thunderbird
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; Geniushits a target no one else can see." Arthur Schopenhauer
Proving the: Simulating termination analyzer Principle
Re: Proving the: Simulating termination analyzer Principle