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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DD specifies non-terminating behavior to HHH --- ONE POINT AT A
TIME !!!
Date: Wed, 19 Feb 2025 18:34:14 -0600
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On 2/19/2025 4:55 AM, Fred. Zwarts wrote:
> Op 18.feb.2025 om 17:48 schreef olcott:
>> On 2/18/2025 8:11 AM, Fred. Zwarts wrote:
>>> Op 18.feb.2025 om 14:37 schreef olcott:
>>>> On 2/18/2025 6:25 AM, Richard Damon wrote:
>>>>> On 2/18/25 6:26 AM, olcott wrote:
>>>>>> On 2/18/2025 3:24 AM, Mikko wrote:
>>>>>>> On 2025-02-17 09:05:42 +0000, Fred. Zwarts said:
>>>>>>>
>>>>>>>> Op 16.feb.2025 om 23:51 schreef olcott:
>>>>>>>>> On 2/16/2025 4:30 PM, joes wrote:
>>>>>>>>>> Am Sun, 16 Feb 2025 15:58:14 -0600 schrieb olcott:
>>>>>>>>>>> On 2/16/2025 2:02 PM, joes wrote:
>>>>>>>>>>>> Am Sun, 16 Feb 2025 13:24:14 -0600 schrieb olcott:
>>>>>>>>>>>>> On 2/16/2025 10:35 AM, joes wrote:
>>>>>>>>>>>>>> Am Sun, 16 Feb 2025 06:51:12 -0600 schrieb olcott:
>>>>>>>>>>>>>>> On 2/15/2025 2:49 AM, Mikko wrote:
>>>>>>>>>>>>>>>> On 2025-02-14 12:40:04 +0000, olcott said:
>>>>>>>>>>>>>>>>> On 2/14/2025 2:58 AM, Mikko wrote:
>>>>>>>>>>>>>>>>>> On 2025-02-14 00:07:23 +0000, olcott said:
>>>>>>>>>>>>>>>>>>> On 2/13/2025 3:20 AM, Mikko wrote:
>>>>>>>>>>>>>>>>>>>> On 2025-02-13 04:21:34 +0000, olcott said:
>>>>>>>>>>>>>>>>>>>>> On 2/12/2025 4:04 AM, Mikko wrote:
>>>>>>>>>>>>>>>>>>>>>> On 2025-02-11 14:41:38 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>>>>>>>> DD correctly simulated by HHH cannot possibly
>>>>>>>>>>>>>>>>> terminate normally.
>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> That claim has already shown to be false. Nothing above
>>>>>>>>>>>>>>>> shows that
>>>>>>>>>>>>>>>> HHH does not return 0. If it does DD also returns 0.
>>>>>>>>>>>>>>> When we are referring to the above DD simulated by HHH
>>>>>>>>>>>>>>> and not
>>>>>>>>>>>>>>> trying to get away with changing the subject to some
>>>>>>>>>>>>>>> other DD
>>>>>>>>>>>>>>> somewhere else
>>>>>>>>>>>>>> such as one that calls a non-aborting version of HHH
>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> then anyone with sufficient knowledge of C programming
>>>>>>>>>>>>>>> knows that no
>>>>>>>>>>>>>>> instance of DD shown above simulated by any corresponding
>>>>>>>>>>>>>>> instance
>>>>>>>>>>>>>>> of HHH can possibly terminate normally.
>>>>>>>>>>>>>> Well, then that corresponding (by what?) HHH isn’t a decider.
>>>>>>>>>>>>> I am focusing on the isomorphic notion of a termination
>>>>>>>>>>>>> analyzer.
>>>>>>>>>>>> (There are other deciders that are not termination analysers.)
>>>>>>>>>>>>
>>>>>>>>>>>>> A simulating termination analyzer correctly rejects any
>>>>>>>>>>>>> input that
>>>>>>>>>>>>> must be aborted to prevent its own non-termination.
>>>>>>>>>>>> Yes, in particular itself is not such an input, because we
>>>>>>>>>>>> *know* that
>>>>>>>>>>>> it halts, because it is a decider. You can’t have your cake
>>>>>>>>>>>> and eat it
>>>>>>>>>>>> too.
>>>>>>>>>>> I am not even using the confusing term "halts".
>>>>>>>>>>> Instead I am using in its place "terminates normally".
>>>>>>>>>>> DD correctly simulated by HHH cannot possibly terminate
>>>>>>>>>>> normally.
>>>>>>>>>> What’s confusing about „halts”? I find it clearer as it does
>>>>>>>>>> not imply
>>>>>>>>>> an ambiguous „abnormal termination”. How does HHH simulate DD
>>>>>>>>>> terminating abnormally, then? Why doesn’t it terminate abnormally
>>>>>>>>>> itself?
>>>>>>>>>> You can substitute the term: the input DD to HHH does not need
>>>>>>>>>> to be
>>>>>>>>>> aborted, because the simulated decider terminates.
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> typedef void (*ptr)();
>>>>>>>>> int HHH(ptr P);
>>>>>>>>>
>>>>>>>>> int DD()
>>>>>>>>> {
>>>>>>>>> int Halt_Status = HHH(DD);
>>>>>>>>> if (Halt_Status)
>>>>>>>>> HERE: goto HERE;
>>>>>>>>> return Halt_Status;
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> int main()
>>>>>>>>> {
>>>>>>>>> HHH(DD);
>>>>>>>>> }
>>>>>>>>>
>>>>>>>>> Every simulated input that must be aborted to
>>>>>>>>> prevent the non-termination of HHH is stipulated
>>>>>>>>> to be correctly rejected by HHH as non-terminating.
>>>>>>>>>
>>>>>>>> A very strange and invalid stipulation.
>>>>>>>
>>>>>>> It merely means that the words do not have their ordinary meaning.
>>>>>>>
>>>>>>
>>>>>> Unless HHH(DD) aborts its simulation of DD itself cannot possibly
>>>>>> terminate normally. Every expert in the C programming language
>>>>>> can see this. People that are not experts get confused by the loop
>>>>>> after the "if" statement.
>>>>>>
>>>>>
>>>>> So? Since it does that, it needs to presume that the copy of itself
>>>>> it sees called does that.
>>>>>
>>>>
>>>> Not at all. Perhaps your technical skill is much more woefully
>>>> deficient than I ever imagined.
>>>>
>>>> Here is the point that you just missed Unless the first HHH
>>>> that sees the non-terminating pattern aborts its simulation
>>>> none of them do because they all have the exact same code.
>>>>
>>>>
>>>
>>> The point Olcott misses is that if the non-terminating HHH is changed
>>> to abort the simulation, the program is changed. He does not
>>> understand that a modification of a program makes a change. Such a
>>> change modifies the behaviour of the program. The non-termination
>>> behaviour has disappeared with this change and only remains in his
>>> dreams. After this change, the simulation would terminate normally
>>> and HHH should no longer abort. But it does, because the code that
>>> detects the 'special condition' has a bug, which makes that it does
>>> not see that the program has been changed into a halting program.
>>
>>
>> When I focus on one single-point:
>> I get two years of dodging and this point is never addressed.
>>
>> [DD simulated by HHH cannot possibly terminate normally]
>>
>> void DDD()
>> {
>> HHH(DDD);
>> return;
>> }
>>
>> int main()
>> {
>> HHH(Infinite_Recursion);
>> HHH(DDD);
>> }
>>
> It is not true that this point has never been addressed. Olcott ignores
> it when it is addressed.
>
> What is the point? Even if HHH fails to simulate the halting program DD
> up to the end because it is logically impossible for it to complete the
> simulation, it still fails.
It fails In the same way that every CAD system
will never correctly represent a geometric circle that has
four equal length sides in the same two dimensional plane.
> If the logically impossible cannot be done,
> we can admit that HHH's simulation fails to complete the impossible task.
> So, why is Olcott trying to fix the logically impossible? He could as
> well try to draw a square circle.
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