Re: Defining a correct simulating halt decider

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.
A correct trace would not only list the instructions, but also the state changes of registers and memory. And it would list the instructions inside HHH when DDD calls HHH.

 It really seems quite ridiculous to me that everyone
could continue to disagree with such easily verified
facts without malevolent motives.

The simulator modified by olcott, however, claims that the same input does not halt. It is clear which of the two is right.
But, without any evidence he disagreed with such easily verified facts without malevolent motives.

 void DDD()
{
   HHH(DDD);
   return;
}
 Is the dumbed down version of the haling problem pathological input:
 int DD(int (*x)())
{
   int Halt_Status = HH(x, x);
   if (Halt_Status)
     HERE: goto HERE;
   return Halt_Status;
}
 I dumbed it down as much as possible and people still don't get
it. They seem to believe that they are free to disagree with the
x86 language.

The input given to the simulator describes a program coded in the x86 language. The semantics of the x86 language allows only one behaviour for the program described by this input. Exact the same input when given for direct execution, for the unmodified famous X86utm simulator and even for HHH1, show that this input describes a halting program.
Olcott has modified the simulator and his version claims that the program does not halt.
Apparently he does not care about the semantics of the x86 language. He just ignores any errors in his code and thinks it is correct to deviate from the semantics of the x86 language in order to make his claims true.
He apparently does not even notice the door that hits his head.
Even with overwhelming evidence, presented by different people with different words, olcott does not get it, because he has brainwashed himself to think that he must be right. Therefore, he cannot understand the proofs of the facts and describes them as lies.