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From: Mikko
Newsgroups: comp.theory
Subject: Re: Everyone here seems to consistently lie about this
Date: Thu, 8 Aug 2024 09:53:08 +0300
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On 2024-08-07 13:14:13 +0000, olcott said:
> 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*
>>>>>>>>>>>
>>>>>>>>>>> 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.
>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> 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.
According to our defamation laws "seemed" is not sufficinet to ensure
that you aren't jailed.
--
Mikko