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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Tue, 3 Sep 2024 14:42:22 -0000 (UTC)
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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.