Sujet : Re: How do computations actually work?
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 23. May 2025, 17:10:19
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <100q6lb$5buc$3@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
User-Agent : Mozilla Thunderbird
On 5/23/2025 2:14 AM, Mikko wrote:
On 2025-05-23 03:31:15 +0000, olcott said:
On 5/22/2025 10:23 PM, wij wrote:
On Thu, 2025-05-22 at 21:47 -0500, olcott wrote:
[cut]
>
Q: How do computations actually work?
A: Computation is merely step-by-step algorithm.
Nothing says it has to be TM.
>
Do the exercises in textbooks first before any claim of it.
>
>
int main()
{
DD(); // by what steps can the HHH that DD calls
} // report on the behavior its caller?
If we don't insist that the report be correct:
1. guess
2. tell what was guessed
This does not work because all computable functions
that implement termination analyzers must compute
the mapping from their input finite string according
to the behavior that it specifies. In my concrete
examples DDD must be simulated by HHH according to the
rules of the x86 language.
_DDD()
[00002172] 55 push ebp ; housekeeping
[00002173] 8bec mov ebp,esp ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404 add esp,+04
[00002182] 5d pop ebp
[00002183] c3 ret
Size in bytes:(0018) [00002183]
DDD specifies that it will continue to call HHH
until HHH sees the repeating pattern and aborts
its simulation.
-- Copyright 2025 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer