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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory,sci.logic
Subject: Re: D correctly simulated by H proved for THREE YEARS --- rewritten
Date: Fri, 14 Jun 2024 07:49:00 -0500
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On 6/14/2024 4:59 AM, Fred. Zwarts wrote:
> Op 13.jun.2024 om 21:41 schreef olcott:
>> On 6/13/2024 2:33 PM, Fred. Zwarts wrote:
>>> Op 13.jun.2024 om 14:44 schreef olcott:
>>>> On 6/13/2024 3:15 AM, Fred. Zwarts wrote:
>>>>> Op 12.jun.2024 om 21:53 schreef olcott:
>>>>>> On 6/12/2024 2:46 PM, Fred. Zwarts wrote:
>>>>>>> Op 12.jun.2024 om 21:20 schreef olcott:
>>>>>>>>
>>>>>>>> On 5/29/2021 2:26 PM, olcott wrote:
>>>>>>>> https://groups.google.com/g/comp.theory/c/dTvIY5NX6b4/m/cHR2ZPgPBAAJ
>>>>>>>>
>>>>>>>> If that was true then you could provide every step of D correctly
>>>>>>>> simulated by H such that D simulated by H reaches its own simulated
>>>>>>>> "ret" instruction.
>>>>>>>
>>>>>>> I said that each H is unable to hit its target, so how could it 
>>>>>>> reach the "ret" instruction of D? Please, think before you reply.
>>>>>>
>>>>>> It is a binary choice either D correctly simulated by H can
>>>>>> possibly terminate normally by reaching its "ret" instruction
>>>>>> or not. Your attempt to twist these words to make it look like
>>>>>> there is more than these two possibilities is either ignorant
>>>>>> or deceptive.
>>>>>>
>>>>>
>>>>> Please, take some more attention to what I said. Read, then think, 
>>>>> before you reply.
>>>>> I said that H is not able to reach its own "ret" when it is 
>>>>> simulating itself. 
>>>>
>>>> That has always been totally irrelevant.
>>>
>>> So, you think that if H does not reach its "ret", D can still reach 
>>> its "ret"?
>>> Try to think. D does not reach its "ret", *because* "H" does not 
>>> reach its "ret".
>>>
>>>>
>>>>> So, no disagreement with that. That proves that H misses its 
>>>>> target. The abort is too early. The target is just some steps 
>>>>> further. It does not mean that the target is at infinity.
>>>>>
>>>>
>>>> The outer H always has one more execution trace to base its halt
>>>> status decision on than any of the nested emulations. This means
>>>> that unless the outer H aborts its simulation then none of them do.
>>>
>>> That is true. But it also means that H aborts one execution trace too 
>>> early. 
>>
>> No it never meant this. 
> 
> Yes, it does mean this. Using another simulator 

Has a different sequence of configurations thus is not
a valid counter-example.

> shows that even the 
> simulated H reaches its "ret". 

I ran the actual code to verify the facts.
HH1(DD,DD) does not have a pathological relationship to its input
thus this input terminates normally.

HH(DD,DD) does have a pathological relationship to its input
thus this input CANNOT POSSIBLY terminate normally.

> It is only that H simulated by itself is 
> aborted too early. Is that so difficult to understand for you?
> 
Aborted too early is false.
Unless HH(DD,DD) aborts pretty soon HH and DD crash due to
out-of-memory error.

>> If H waits for some other H to abort their
>> simulation then H waits forever.
> 
> There is no other H.  

Clearly you hardly understand anything that I have been saying.
(a) HH(DD,DD) directly executed in main simulates its input.
(b) The simulated DD calls a simulated HH(DD,DD) that
(c) simulates another instance of DD... goto (b)

HH aborts as soon as it can after seeing DD repeat all of
itself states exactly once. If HH waited for fifteen cycles
(and did not run out of memory) it would still see one
more cycle than the next inner HH.

Either the outermost HH aborts or none of them do.

>This H aborts too early. This H does not wait, so 
> it does not help to dream of another H that waits. H does what it is 
> programmed to do and aborts too early, because that is the fundamental 
> problem of a simulator simulating itself. It will never see its final 
> simulated state.
> 
>>  H is always at least one execution
>> trace ahead of every other H.
> 
> Exactly! That is the reason why the abort one execution trace too early. 
> It seems you start to see it. H will never see that it is only some 
> steps from the final state of its simulation, because it aborts before 
> it can see that.
> That does not mean that there is an infinitely repeated recursion, but 
> that the recursion is only repeated one time more than can be simulated 
> by H. That is the fundamental problem of a simulator simulating itself.
> 
> You can try to simulate longer, but that does not help. The simulation 
> invariant is that the abort is always one execution trace too early. The 
> other invariant is that in an aborting simulator there is never an 
> infinitely repeated recursion.

-- 
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer