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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Tue, 3 Sep 2024 22:16:59 -0400
Organization: i2pn2 (i2pn.org)
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On 9/3/24 2:40 PM, olcott wrote:
> On 9/3/2024 9:42 AM, joes wrote:
>> 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.
> 
> DDD emulated by HHH cannot possibly reach
> its final halt state no matter what HHH does.

Of course it does, because the HHH that it calls aborts and returns to it.

You are confusing the PROGRAM DDD, which happened to be emulated by HHH, 
with the emulation of DDD by HHH, which is a different thing.


This apparently INTENTIONAL confusion (since you don't correct yourself 
after the error is exposed) just shows that you are being INTENTIONALLY 
deceptive and a pathological liar.

> 
>>>>> A halt decider never ever computes the mapping for the computation
>>>>> that itself is contained within.
>> Then it is not total.
> 
> Yes it is you are wrong.

No. Just proves you are a stupid liar.

> 
>>>>> 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.
> 
> Which makes the pathological input a counter-example
> to the false assumption that the direct execution of
> a machine always has the same behavior as the machine
> simulated by its pathological simulator.

Nope, since you can't point out which instruction was CORRECTLY emulated 
that differed in behavior.

All you have shown is that HHH, if it DOES correctly emulate the input 
(meaning the call HHH instructions go into HHH) then all it sees is 
exactly the same execution trace as the directly executed DDD (unless 
HHH isn't a pure function) until it gives up an aborts.

That does NOT show that the "correct simulation by HHH" is different 
then the direct execution, it shows that HHH makes an incorrect 
deduction about the behavior when it decides to abort and the "guess" 
based on lies what the behavior will be.

> 
>>>>> 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?
>>
> 
> void DDD()
> {
>    HHH(DDD);
>    OutputString("This code is unreachable by DDD emulated by HHH");
> }

But DDD when run will get there, as will DDD correctly and fully 
emulated by HHH1.

> 
>>>> 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.
>>
> 
> No you are wrong !!!
> 

No, you are LYING because it seems you don't even KNOW what a CORRECT 
emulation actualy is, or what it means by the behavior of DDD.