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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: H(D,D) cannot even be asked about the behavior of D(D) V3
 ---IGNORING ALL OTHER REPLIES
Date: Wed, 19 Jun 2024 12:07:05 -0500
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On 6/19/2024 11:29 AM, Alan Mackenzie wrote:
> olcott <polcott333@gmail.com> wrote:
>> On 6/19/2024 10:39 AM, Fred. Zwarts wrote:
>>> Op 19.jun.2024 om 16:55 schreef olcott:
>>>> On 6/19/2024 8:46 AM, Fred. Zwarts wrote:
> 
> [ .... ]
> 
>>>>> It seems really to difficult for you. So, you prefer to forget or
>>>>> ignore it.
>>>>> The 'call' instruction at 000020aa is incorrectly simulated.
> 
>>>> As a matter of fact it is not incorrectly simulated.
>>>> I am showing this with HH0 instead of H0 because
>>>> the trace provided by HH0 is easier to understand.
> 
> So instead of why you think that instruction is correctly simulated, you
> just spam the newsgroup with more barely penetrable machine code.
> 
>>>>> H0 is required to halt, i.e. to return, but your simulation does not
>>>>> show the 'ret' instruction of H0.
> 
>>>> Yes it does not show this yet HH0 does simulate itself simulating DDD.
> 
>>>> If the eight lines of DDD correctly simulated by HH0 were mixed in
>>>> with the 150 pages of HH0 simulating itself it would be too difficult
>>>> to see the behavior of DDD. The reader would have to carefully search
>>>> for the machine addresses of DDD that only occur every 19 pages.
> 
> Of course, you could always have found these places yourself and included
> them in a post, thus making it plausible that these places actually
> exist, and that you have done such an execution trace.
> 

The best way to do this is to make a compiler flag that enables
both versions to be output. After more than three years not one
person could correctly understand the single page of x86 code
so I am sure that 150 more pages are not going to make this easier
for them.

>>>> I will adapt HH0 so that it does show HH0 simulating itself
>>>> simulating DDD.
> 
>>>>> It seems you are so confused that you do not understand it.
>>>>> Therefore, you think it is a change of subject or gibberish. Showing
>>>>> that it is over your head.
>>>>> Instead of fixing the problem, you just repeat the claim without any
>>>>> new argument.
>>>>> The simulation fails, because it is aborted one cycle too soon,
>>>>> before the simulated H0 would reach its 'ret' instruction. A correct
>>>>> simulation would see this. Unfortunately, H0 is unable to correctly
>>>>> simulate itself.
> 
>>>> Because the executed HH0 always has at least one more execution trace
>>>> than any of its simulated instances unless it aborts the simulation
>>>> after a fixed number of repeating states, none of them do.
> 
>>> What is your problem:
>>> 1) You say: 'HH0 always has at least one more execution trace than any
>>> of its simulated instances'. That is correct. That is true even if it
>>> does abort. Is that over your head?
>>> 2) If it aborts, it misses the fact that the simulation of itself would
>>> also abort one execution trace later. Is that over your head?
>>> 3) If it misses the fact that its simulation of itself would halt one
>>> execution trace later, it is incorrect to report non-halting. Is that
>>> over your head?
> 
>>> What is your problem: 1, 2, or 3.
>>> If you do not even understand it, you better stop talking about it.
> 
>> *THIS IS SIMPLY OVER YOUR HEAD*
>> *THIS IS SIMPLY OVER YOUR HEAD*
>> *THIS IS SIMPLY OVER YOUR HEAD*
> 
> [ more spam deleted. ]
> 
>> Every C programmer that knows what an x86 emulator is knows that when
>> H0 emulates the machine language of Infinite_Loop, Infinite_Recursion,
>> and DDD that it must abort these emulations so that itself can terminate
>> normally.
> 
>> Every C programmer has agreed thus you simply don't know these things
>> well enough.
> 

All of the C programmers that have stated an opinion (at least six)
have agreed with me.

> That is another of your lies.  Every C programmer has NOT agreed this.
> In particular, I haven't.
> 

All of the C programmers on the C programming groups have
agreed with me.

> Fred knows these things extremely well.  It is you that appears to have a
> problem with them.
> 

*If that was actually true then you would stop dodging this challenge*
It is dishonest for you to erase this challenge and call it spam:

void DDD()
{
   H0(DDD);
}

_DDD()
[000020a2] 55         push ebp      ; housekeeping
[000020a3] 8bec       mov ebp,esp   ; housekeeping
[000020a5] 68a2200000 push 000020a2 ; push DDD
[000020aa] e8f3f9ffff call 00001aa2 ; call H0
[000020af] 83c404     add esp,+04   ; housekeeping
[000020b2] 5d         pop ebp       ; housekeeping
[000020b3] c3         ret           ; never gets here
Size in bytes:(0018) [000020b3]

Exactly which step of DDD emulated by H0 was emulated
incorrectly such that this emulation would be complete?
AKA DDD emulated by H0 reaches machine address [000020b3]


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

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