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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: HHH computes the mapping from its input finite sting to the
 actual behavior specified by this finite string (Which is the results of
 running the input)
Date: Wed, 7 Aug 2024 21:03:11 -0400
Organization: i2pn2 (i2pn.org)
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On 8/7/24 9:54 AM, olcott wrote:
> On 8/7/2024 2:29 AM, Mikko wrote:
>> On 2024-08-05 13:49:44 +0000, olcott said:
>>
>>> On 8/5/2024 2:39 AM, Mikko wrote:
>>>> On 2024-08-04 18:59:03 +0000, olcott said:
>>>>
>>>>> On 8/4/2024 1:51 PM, Richard Damon wrote:
>>>>>> On 8/4/24 9:53 AM, olcott wrote:
>>>>>>> On 8/4/2024 1:22 AM, Fred. Zwarts wrote:
>>>>>>>> Op 03.aug.2024 om 18:35 schreef olcott:
>>>>>>>  >>>> ∞ instructions of DDD correctly emulated by HHH[∞] never
>>>>>>>>> reach their own "return" instruction final state.
>>>>>>>>>
>>>>>>>>> So you are saying that the infinite one does?
>>>>>>>>>
>>>>>>>>
>>>>>>>> Dreaming again of HHH that does not abort? Dreams are no 
>>>>>>>> substitute for facts.
>>>>>>>> The HHH that aborts and halts, halts. A tautology.
>>>>>>>
>>>>>>> void DDD()
>>>>>>> {
>>>>>>>    HHH(DDD);
>>>>>>>    return;
>>>>>>> }
>>>>>>>
>>>>>>> That is the right answer to the wrong question.
>>>>>>> I am asking whether or not DDD emulated by HHH
>>>>>>> reaches its "return" instruction.
>>>>>>
>>>>>> But the "DDD emulated by HHH" is the program DDD above,
>>>>>
>>>>> When I say DDD emulated by HHH I mean at any level of
>>>>> emulation and not and direct execution.
>>>>
>>>> If you mean anything other than what the words mean you wihout
>>>> a definition in the beginning of the same message then it is
>>>> not reasonable to expect anyone to understand what you mean.
>>>> Instead people may think that you mean what you say or that
>>>> you don't know what you are saying.
>>>
>>> If you don't understand what the word "emulate" means look it up.
>>
>> I know what it means. But the inflected form "emulated" does not
>> mean what you apparently think it means. You seem to think that
>> "DDD emulated by HHH" means whatever HHH thinks DDD means but it
>> does not. DDD means what it means whether HHH emulates it or not.
>>
> 
> In other words when DDD is defined to have a pathological
> relationship to HHH we can just close our eyes and ignore
> it and pretend that it doesn't exist?

No, you have to admit that you will get the wrong answer.

Remember, HHH is a fixed defined piece of code, so it has a fixed 
defined answer that it WILL give based on the algorithm that HHH used.

What ever that answer is, it will be wrong, as can be easily determined 
by many other deciders.

> 
> DDD does specify non-halting behavior to HHH and HHH must
> report on this non-halting behavior that DDD specifies.

No, because halting is an OBJECTIVE criteria, and either HHH will return 
and that makes DDD a halting program (so HHH would have needed to have 
guessed 1 to be right), or HHH will never return, making DDD a 
non-halting program (and ONLY that makes it non-halting) but HHH turns 
out not to be a decider.


> 
> All halt deciders compute the mapping from their input
> finite string to the behavior that this finite string
> specifies.

Which, for a halt decide is DEFINED to be the mapping based on the 
direct execution of the program, and whether it reaches a final state 
iin a finite number of steps or not.

That HHH can't compute this does not make that not the mapping it NEEDS 
to compute to be a Halt Decider, that just means there is no such thing 
as a universally correct Halt Decider.

> 
> No halt decider is ever allowed to report on the behavior
> of any computation that itself is contained within unless
> this is the same behavior that its finite string input
> specifies.

Says WHAT? This is just you stating you lie again.

ALL Halt deciders are REQUIRED (to be a halt decider) that answer about 
*ANY* program described by their input, even if it includes a copy of 
itself.

You just proved yourself to be a ingnoran LIAR that reckless repeats 
lies that he has been informed are lies because you stupidly refuse to 
even look at the actual definition, because you don't "beleive" in them.

Sorry, you don't get to not believe in the rules of the game that have 
been establish.

You just canceled yourself out of the game, and booked your ticket to 
the garbage heap of life.

> 
> void DDD()
> {
>    HHH(DDD);
>    return;
> }
> 
> Any expert in the C language that knows what x86 emulators
> are knows that DDD correctly emulated by HHH specifies what
> is essentially equivalent to infinite recursion that cannot
> possibly reach its "return" instruction halt state.
> 
> It seems that no one here has that degree of expertise.
> That they know that they don't understand these things
> and still say that I am wrong is dishonest.
>