Sujet : Re: DDD correctly emulated by HHH cannot possibly halt
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 10. Jul 2024, 15:25:54
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v6m253$1tj30$4@dont-email.me>
References : 1 2
User-Agent : Mozilla Thunderbird
On 7/10/2024 2:02 AM, Mikko wrote:
On 2024-07-09 23:49:16 +0000, olcott said:
_DDD()
[00002163] 55 push ebp ; housekeeping
[00002164] 8bec mov ebp,esp ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
>
DDD correctly emulated by any pure function HHH that
correctly emulates 1 to ∞ steps of DDD can't make it
past the above line of code no matter what.
>
[00002170] 83c404 add esp,+04
[00002173] 5d pop ebp
[00002174] c3 ret
Size in bytes:(0018) [00002174]
The subject line is misleading. There is only one DDD so "DDD correctly
emulated by HHH" should simply mean DDD and nothing else.
*I added this to my latest paper*
Every time any HHH correctly emulates DDD it calls the
x86utm operating system to create a separate process
context with its own memory virtual registers and stack,
thus each recursively emulated DDD is a different instance.
Simulating Termination Analyzer H is Not Fooled by Pathological Input D
https://www.researchgate.net/publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D --
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
…