Re: Defining a correct simulating halt decider

On 9/9/24 1:40 PM, olcott wrote:

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? I told you the question too many times and
you always dishonestly change it.
 

No, it has always been, when HHH does what it does (which is abort) does the program DDD that calls that HHH reach its final state or not.
It has NEVER been about can HHH emulate its input to a final state, like you keep on wanting to talk about, as you dont seem to understand that the partial emulation by HHH doesn't actual define the behavior of the input, only what HHH knows about the behavior of the input.