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From: olcott <abc@def.com>
Newsgroups: comp.theory
Subject: Re: Everyone here seems to consistently lie about this
Date: Wed, 7 Aug 2024 08:14:13 -0500
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On 8/7/2024 2:12 AM, Mikko wrote:
> On 2024-08-06 11:35:51 +0000, olcott said:
> 
>> On 8/6/2024 3:07 AM, Mikko wrote:
>>> On 2024-08-05 12:45:11 +0000, olcott said:
>>>
>>>> On 8/5/2024 2:27 AM, Mikko wrote:
>>>>> On 2024-08-04 12:33:20 +0000, olcott said:
>>>>>
>>>>>> On 8/4/2024 2:15 AM, Mikko wrote:
>>>>>>> On 2024-08-03 13:48:12 +0000, olcott said:
>>>>>>>
>>>>>>>> On 8/3/2024 3:06 AM, Mikko wrote:
>>>>>>>>> On 2024-08-02 02:09:38 +0000, olcott said:
>>>>>>>>>
>>>>>>>>>> *This algorithm is used by all the simulating termination 
>>>>>>>>>> analyzers*
>>>>>>>>>> <MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>>>> 10/13/2022>
>>>>>>>>>>      If simulating halt decider H correctly simulates its input D
>>>>>>>>>>      until H correctly determines that its simulated D would 
>>>>>>>>>> never
>>>>>>>>>>      stop running unless aborted then
>>>>>>>>>>
>>>>>>>>>>      H can abort its simulation of D and correctly report that D
>>>>>>>>>>      specifies a non-halting sequence of configurations.
>>>>>>>>>> </MIT Professor Sipser agreed to ONLY these verbatim words 
>>>>>>>>>> 10/13/2022>
>>>>>>>>>>
>>>>>>>>>> DDD is correctly emulated by HHH according to the x86
>>>>>>>>>> language semantics of DDD and HHH including when DDD
>>>>>>>>>> emulates itself emulating DDD
>>>>>>>>>>
>>>>>>>>>> *UNTIL*
>>>>>>>>>>
>>>>>>>>>> HHH correctly determines that never aborting this
>>>>>>>>>> emulation would cause DDD and HHH to endlessly repeat.
>>>>>>>>>
>>>>>>>>> The determination is not correct. DDD is a halting computation, as
>>>>>>>>> correctely determined by HHH1 or simly calling it from main. It is
>>>>>>>>> not possible to correctly determine that ha haling computation is
>>>>>>>>> non-halting, as is self-evdent from the meaning of the words.
>>>>>>>>>
>>>>>>>>
>>>>>>>> [Who here is too stupid to know that DDD correctly simulated
>>>>>>>>   by HHH cannot possibly reach its own return instruction?]
>>>>>>>
>>>>>>> Who here is too stupid to know that whether DDD can reach its
>>>>>>> own return instruction depends on code not shown below?
>>>>>>>
>>>>>>
>>>>>> void DDD()
>>>>>> {
>>>>>>    HHH(DDD);
>>>>>>    return;
>>>>>> }
>>>>>>
>>>>>> It is stipulated that HHH is an x86 emulator the emulates
>>>>>> N instructions of DDD where N is 0 to infinity.
>>>>>
>>>>> That is not stipulated above. Anyway, that stipulation would not
>>>>> alter the correctness of my answer.
>>>>>
>>>>
>>>> typedef void (*ptr)();
>>>> int HHH(ptr P);
>>>>
>>>> void DDD()
>>>> {
>>>>    HHH(DDD);
>>>>    return;
>>>> }
>>>>
>>>> int main()
>>>> {
>>>>    HHH(DDD);
>>>> }
>>>>
>>>> In other words you do not know C well enough to comprehend
>>>> that DDD correctly simulated by any HHH cannot possibly reach
>>>> its own "return" instruction halt state.
>>>
>>> You are lying again.
>>>
>>
>> I am hypothesizing.
> 
> You were and still are lying. There was no word (such as "assume") in
> your calim about me, so it was not a hypothesis but a lie.
> 

Several of your answers seemed to show that you did not
know C very well. Fred and Joes did not seem to know
programming very well.

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