Re: Defining a correct halt decider

Op 03.sep.2024 om 21:54 schreef olcott:

On 9/3/2024 1:53 PM, joes wrote:

Am Tue, 03 Sep 2024 08:17:56 -0500 schrieb olcott:

On 9/3/2024 3:44 AM, Mikko wrote:

On 2024-09-02 16:06:11 +0000, olcott said:
>

A correct halt decider is a Turing machine T with one accept state and
one reject state such that:
If T is executed with initial tape contents equal to an encoding of
Turing machine X and its initial tape contents Y, and execution of a
real machine X with initial tape contents Y eventually halts, the
execution of T eventually ends up in the accept state and then stops.
If T is executed with initial tape contents equal to an encoding of
Turing machine X and its initial tape contents Y, and execution of a
real machine X with initial tape contents Y does not eventually halt,
the execution of T eventually ends up in the reject state and then
stops.

Your "definition" fails to specify "encoding". There is no standard
encoding of Turing machines and tape contents.
>

That is why I made the isomorphic x86utm system.
By failing to have such a concrete system all kinds of false assumptions
cannot be refuted.

What would those assumptions be?
>

The behavior of DDD emulated by HHH** <is> different than the behavior
of the directly executed DDD** **according to the semantics of the x86
language

How can the same code have different semantics?
>

 The pathological relationship between DDD and HHH really
cannot be simply ignored as if it does not exist.
 

HHH is required to report on the behavior tat its finite string input
specifies even when this requires HHH to emulate itself emulating DDD.

 > The input specifies an aborting HHH - which you don’t simulate.>
 void DDD()
{
   HHH(DDD);
   OutputString("This code is unreachable by DDD emulated by HHH");
}
 

DDD never halts unless it reaches its own final halt state. The fact
that the executed HHH halts has nothing to do with this.

Other than that DDD calls HHH?
>

HHH is not allowed to report on the computation that itself is contained
within.

Then it is only partial, and doesn’t even solve the case it was built for.
>

 int sum(int x, int y);
sum(3,4) is not allowed to report on the sum of 5 + 6
for the same reason. HHH(DDD) cannot report on behavior
that it cannot see.

Exactly, so it should not report on halting behaviour if its stops the simulation before the simulation could halt.
If the simulator prevents the simulation to halt, then there is no reason to report about the halting behaviour.

 HHH cannot correctly report on the AFTER-THE-FACT
behavior that it has aborted its simulation BEFORE-THE-FACT.

Olcott seems to forget that the action to abort was programmed already BEFORE the abort took place. So, the simulated HHH was programmed to see the 'special condition' and abort, BEFORE the simulating HHH aborted it.
It is incorrect to assume that the abort code that is present BEFORE the abort takes place, would not be executed in the simulated program, only because the simulation has not yet reached it.
In other words, before-the-fact of the abort, the simulation did not halt and there is no reason to decide that there is non-halting behaviour. So, the only reason to decide about the halting behaviour is found in the code that would be executed after-the-fact. And there we find the code to see the detection of a 'special condition' and the abort.

 

Except for the case of pathological self-reference the behavior of the
directly executed machine M is always the same as the correctly
simulated finite string ⟨M⟩.

 

That sure sounds like a mistake to me.
>

 THE EXECUTION TRACE HAS ALWAYS PROVED THAT I AM CORRECT
FOR THREE FREAKING YEARS all the way back when it was
P correctly emulated by D.

No, it has always shown that the abort was too soon, one cycle before the simulated HHH would reach the code to see the 'special condition', after which it would abort and return.

 IT REMAINS A VERIFIED FACT THAT DDD EMULATED BY HHH CANNOT
POSSIBLY REACH ITS OWN FINAL HALT STATE,

Exactly, I have repeated this many times, because it proves that the simulation is incorrect.
HHH cannot possibly simulate itself correctly up to the end. That there is an end is proved by the direct executions, by the simulations of the world class simulator and even by HHH1.
The meaning of the input, a finite string that describes the program, is fixed by the semantics of the x86 language. It does not depend on who or what interprets it.
But olcott thinks he can change the meaning of it by using a crippled simulator that is unable to reach the end of the simulation.