Re: Defining a correct simulating halt decider

Am Mon, 02 Sep 2024 16:06:24 -0500 schrieb olcott:

On 9/2/2024 12:52 PM, Fred. Zwarts wrote:

Op 02.sep.2024 om 18:38 schreef olcott:

A halt decider is a Turing machine that computes the mapping from its
finite string input to the behavior that this finite string specifies.
If the finite string machine string machine description specifies that
it cannot possibly reach its own final halt state then this machine
description specifies non-halting behavior.

Which DDD does not.

A halt decider never ever computes the mapping for the computation
that itself is contained within.

Then it is not total.

Unless there is a pathological relationship between the halt decider H
and its input D the direct execution of this input D will always have
identical behavior to D correctly simulated by simulating halt decider
H.

Which makes this pathological input a counterexample.

A correct emulation of DDD by HHH only requires that HHH emulate the
instructions of DDD** including when DDD calls HHH in recursive
emulation such that HHH emulates itself emulating DDD.

Indeed, it should simulate *itself* and not a hypothetical other HHH
with different behaviour.

It is emulating the exact same freaking machine code that the x86utm
operating system is emulating.

It is not simulating the abort because of a static variable. Why?

If HHH includes code to see a 'special condition' and aborts and halts,
then it should also simulate the HHH that includes this same code and

DDD has itself and the emulated HHH stuck in recursive emulation.

Your HHH incorrectly changes behaviour.

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.