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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: DD specifies non-terminating behavior to HHH --- ONE POINT AT A
 TIME !!!
Date: Wed, 19 Feb 2025 11:55:45 +0100
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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. 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.