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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Any honest person that knows the x86 language can see... predict
 correctly
Date: Thu, 1 Aug 2024 19:39:53 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Thu, 01 Aug 2024 10:50:29 -0500 schrieb olcott:
> 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.
When one introduces it, it becomes unnecessary, because the cycle stops
from the INSIDE.
After how many repetitions can infinite recursion be aborted before
the simulation becomes wrong?

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.