Re: Defining a correct simulating halt decider

Op 09.sep.2024 om 19:40 schreef olcott:

On 9/9/2024 5:57 AM, Fred. Zwarts wrote:

Op 08.sep.2024 om 15:10 schreef olcott:

On 9/8/2024 7:46 AM, joes wrote:

Am Sat, 07 Sep 2024 08:56:02 -0500 schrieb olcott:

On 9/7/2024 3:27 AM, Mikko wrote:

On 2024-09-06 11:42:48 +0000, olcott said:

On 9/6/2024 6:19 AM, Mikko wrote:

On 2024-09-05 13:24:20 +0000, olcott said:

On 9/5/2024 2:34 AM, Mikko wrote:

On 2024-09-03 13:00:50 +0000, olcott said:

On 9/3/2024 5:25 AM, Mikko wrote:

On 2024-09-02 16:38:03 +0000, olcott said:
>

A halt decider is a Turing machine that computes the mapping
from its finite string input to the behavior that this finite
string specifies.

A halt decider needn't compute the full behaviour, only whether
that behaviour is finite or infinite.

Like Sipser said.
>

New slave_stack at:1038c4 Begin Local Halt Decider Simulation
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
Hence  HHH(DDD)==0 is correct

>
Nice to see that you don't disagree with what said.
Unvortunately I can't agree with what you say.
HHH terminates, so DDD obviously terminates, too.

>
DDD emulated by HHH never reaches it final halt state.

If that iis true it means that HHH called by DDD does not return and
therefore is not a ceicder.

The directly executed HHH is a decider.

>
If the called HHH behaves differently from the direcly executed HHH
then the DDD is not relevant to classic proofs of the impossibility of
a halting decider.
If you can't show encoding rules that permit the encoidng of the
behaviour of the directly executed DDD to HHH then HHH is not a halting
decider.

I SHOW THE ACTUAL EXECUTION TRACE AND EVERYONE DISAGREES WITH IT.

Your implementation is buggy.
>

>
X86utm is based on a world class x86 emulator that
has had decades of development effort. It has been
trivial to verify to the execution traces that it
produces are correct for three years.

>
And the simulation by this unmodified X86utm showed that the DDD based on the HHH that aborts, halts.
>

 The freaking question has never been when DDD is aborted does
it stop running?

This shows that olcott's language processing is too bad to even understand what the question is.
The question is: does the finite string which describes the program based on the DDD that uses the HHH that aborts, describe a halting program?
The answer to this question is: Yes this program halts. It is proven by the direct execution, by the simulation of the unmodified world class simulator, even by HHH1, but HHH decides the opposite.
The semantics of the x86 language allows only one behaviour of this finite string. This proves that HHH is incorrect and violates the semantics of the x86 language.
HHH cannot possibly simulate itself correctly up to the end.