Sujet : Re: Any honest person that knows the x86 language can see... predict correctly
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 01. Aug 2024, 16:50:29
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v8gas5$284fm$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
User-Agent : Mozilla Thunderbird
On 8/1/2024 10:44 AM, joes wrote:
Am Thu, 01 Aug 2024 08:04:23 -0500 schrieb olcott:
On 8/1/2024 7:56 AM, Fred. Zwarts wrote:
Op 01.aug.2024 om 13:51 schreef olcott:
On 8/1/2024 2:46 AM, Fred. Zwarts wrote:
Op 01.aug.2024 om 05:51 schreef olcott:
On 7/31/2024 10:08 PM, wij wrote:
On Tue, 2024-07-30 at 18:50 -0500, olcott wrote:
>
It is not supposed to be a general solution to the halting
problem.
it only shows how the "impossible" input is correctly determined
to be non halting.
But how do you determine it is non-halting?
As I know you are even unable to define what 'halt' mean !!!
If DDD correctly emulated by HHH cannot possibly reach its return
instruction then it never halts.
>
But a correct simulation is impossible.
When HHH does what-ever-the-hell the x86 semantics specifies then HHH
is correct.
But since HHH deviates from the semantics of the x86 language (by
skipping instructions of a halting program) it is incorrect.
>
Only a freaking moron would believe that a non terminating input should
be simulated forever.
I mean, how many iterations of an infinite loop can I skip simulating or
how many do I have to simulate to get identical behaviour?
When one disables the abort code then the cycle never stops.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
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