Re: Simulating termination analyzers by dummies --- criteria is met

Am Sat, 22 Jun 2024 09:11:28 -0500 schrieb olcott:

On 6/22/2024 8:27 AM, Richard Damon wrote:

On 6/22/24 9:04 AM, olcott wrote:

On 6/22/2024 3:05 AM, Mikko wrote:

On 2024-06-21 13:19:28 +0000, olcott said:

On 6/21/2024 2:11 AM, Mikko wrote:

On 2024-06-20 15:23:09 +0000, olcott said:

On 6/20/2024 10:08 AM, Mikko wrote:

On 2024-06-20 05:40:28 +0000, olcott said:

On 6/20/2024 12:29 AM, Mikko wrote:

On 2024-06-19 14:05:29 +0000, olcott said:

On 6/19/2024 4:29 AM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 6/18/2024 4:36 PM, Alan Mackenzie wrote:

In comp.theory olcott <polcott333@gmail.com> wrote:

On 6/18/2024 12:57 PM, joes wrote:

Am Tue, 18 Jun 2024 12:25:44 -0500 schrieb olcott:

On 6/18/2024 12:06 PM, joes wrote:

I use C because there are zero gaps in exactly what it means.

>
THere are lont of gaps in C. Some are mistakes that are corrected
in technical corrigenda. Others are undefined and implementation
defined behaviour. Your program uses non-standard extensions to C
so it does not communicate well. If also is too big to be a part
of a publishable article.
>

*There are zero gaps in the behavior of DDD correctly simulated by
HH0*

That is a completely different statement. Still wrong, as HH0 can't
simulate itself.

Linz Ĥ is fully defined in terms of H, so its behaviour can be
inferred from the behaviour of H. Therefore Linz can prove about
the behaviour of both Ĥ and H what needs be proven.

Ben Bacarisse contacted professor Sipser to verify that he really did
says this. The details are in this forum about the same date.

And, as I remember, he also verified that he disagrees with your
definition of correct simulation.

On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game. PO really /has/ an H (it's
 > trivial to do for this one case) that correctly determines that
 > P(P) *would* never stop running *unless* aborted.

Right, Ben was willing to do what I am not that you can prove that, by
your definition, H can show that it "must" abort its simulation or the
input will run forever.
But, just like me, he also agrees that this is NOT the defintion of
Halting, so H is just shown to be a correct (partial) POOP decider but
ot a Halt Decider, not even for that one input.

On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > He knows and accepts that P(P) actually does stop. The wrong answer
 > is justified by what would happen if H (and hence a different P)
 > were not what they actually are.
*Ben agrees that the criteria is met for the input*
*Ben disagrees that the criteria is met for the non-input*
Yet no one here can stay focused on the fact that non-inputs *DO NOT
COUNT*

The input is always just DDD, along with its embedded "decider".

void DDD()
{
   HHH0(DDD);
}
int main()
{
   Output("Input_Halts = ", HHH0(DDD));
   Output("Input_Halts = ", HHH1(DDD));
}
 
It is a verified fact that the behavior that finite string DDD presents
to HH0 is that when DDD correctly simulated by HH0 calls HH0(DDD) that
this call DOES NOT RETURN.

Making DDD's halting undecidable.

It is a verified fact that the behavior that finite string DDD presents
to HH1 is that when DDD correctly simulated by HH1 calls HH0(DDD) that
this call DOES RETURN.

This is the correct behaviour.

--
Man kann mit dunklen Zahlen nicht rechnen. Für die eigentliche Mathematik
sind sie vollkommen nutzlos. --Wolfgang Mückenheim