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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: How do computations actually work?
Date: Mon, 26 May 2025 16:50:49 -0400
Organization: i2pn2 (i2pn.org)
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On 5/26/25 12:26 PM, olcott wrote:
> On 5/26/2025 11:05 AM, Fred. Zwarts wrote:
>> Op 26.mei.2025 om 17:50 schreef olcott:
>>> On 5/26/2025 3:38 AM, Mikko wrote:
>>>> On 2025-05-25 14:50:58 +0000, olcott said:
>>>>
>>>>> On 5/25/2025 4:09 AM, Mikko wrote:
>>>>>> On 2025-05-24 15:25:21 +0000, olcott said:
>>>>>>
>>>>>>> On 5/24/2025 2:54 AM, Mikko wrote:
>>>>>>>> On 2025-05-23 16:04:49 +0000, olcott said:
>>>>>>>>
>>>>>>>>> On 5/23/2025 2:09 AM, Mikko wrote:
>>>>>>>>>> On 2025-05-23 02:47:40 +0000, olcott said:
>>>>>>>>>>
>>>>>>>>>>> On 5/22/2025 8:24 PM, Mike Terry wrote:
>>>>>>>>>>>> On 22/05/2025 06:41, Richard Heathfield wrote:
>>>>>>>>>>>>> On 22/05/2025 06:23, Keith Thompson wrote:
>>>>>>>>>>>>>> Richard Heathfield <rjh@cpax.org.uk> writes:
>>>>>>>>>>>>>>> On 22/05/2025 00:14, olcott wrote:
>>>>>>>>>>>>>>>> On 5/21/2025 6:11 PM, Richard Heathfield wrote:
>>>>>>>>>>>>>> [...]
>>>>>>>>>>>>>>>>> Turing proved that what you're asking is impossible.
>>>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>>> That is not what he proved.
>>>>>>>>>>>>>>>
>>>>>>>>>>>>>>> Then you'll be able to write a universal termination 
>>>>>>>>>>>>>>> analyser that can
>>>>>>>>>>>>>>> correctly report for any program and any input whether it 
>>>>>>>>>>>>>>> halts. Good
>>>>>>>>>>>>>>> luck with that.
>>>>>>>>>>>>>>
>>>>>>>>>>>>>> Not necessarily.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Of course not. But I'm just reflecting. He seemed to think 
>>>>>>>>>>>>> that my inability to write the kind of program Turing 
>>>>>>>>>>>>> envisaged (an inability that I readily concede) is evidence 
>>>>>>>>>>>>> for his argument. Well, what's sauce for the goose is sauce 
>>>>>>>>>>>>> for the gander.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> Even if olcott had refuted the proofs of the
>>>>>>>>>>>>>> insolvability of the Halting Problem -- or even if he had 
>>>>>>>>>>>>>> proved
>>>>>>>>>>>>>> that a universal halt decider is possible
>>>>>>>>>>>>>
>>>>>>>>>>>>> And we both know what we both think of that idea.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> -- that doesn't imply
>>>>>>>>>>>>>> that he or anyone else would be able to write one.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Indeed.
>>>>>>>>>>>>>
>>>>>>>>>>>>>> I've never been entirely clear on what olcott is claiming.
>>>>>>>>>>>>>
>>>>>>>>>>>>> Nor I. Mike Terry seems to have a pretty good handle on it, 
>>>>>>>>>>>>> but no matter how clearly he explains it to me my eyes 
>>>>>>>>>>>>> glaze over and I start to snore.
>>>>>>>>>>>>
>>>>>>>>>>>> Hey, it's the way I tell 'em!
>>>>>>>>>>>>
>>>>>>>>>>>> Here's what the tabloids might have said about it, if it had 
>>>>>>>>>>>> made the front pages when the story broke:
>>>>>>>>>>>>
>>>>>>>>>>>>   COMPUTER BOFFIN IS TURING IN HIS GRAVE!
>>>>>>>>>>>>
>>>>>>>>>>>>   An Internet crank claims to have refuted Linz HP proof by 
>>>>>>>>>>>> creating a
>>>>>>>>>>>>   Halt Decider that CORRECTLY decides its own "impossible 
>>>>>>>>>>>> input"!
>>>>>>>>>>>>   The computing world is underwhelmed.
>>>>>>>>>>>>
>>>>>>>>>>>> Better?  (Appologies for the headline, it's the best I could 
>>>>>>>>>>>> come up with.)
>>>>>>>>>>>>
>>>>>>>>>>>> Mike.
>>>>>>>>>>>>
>>>>>>>>>>>
>>>>>>>>>>> There is a key detail about ALL of these proofs
>>>>>>>>>>> that no one has paid attention to for 90 years.
>>>>>>>>>>>
>>>>>>>>>>> It is impossible to define *AN INPUT* to HHH that
>>>>>>>>>>> does the opposite of whatever value that HHH returns.
>>>>>>>>>>
>>>>>>>>>> That is a key detail about HHH. Your HHH is not a part of 
>>>>>>>>>> those proofs.
>>>>>>>>>
>>>>>>>>> All of the proofs work this same way.
>>>>>>>>
>>>>>>>> No, they don't. Some proofs derive the same conclusion with an 
>>>>>>>> essentially
>>>>>>>> different approach.
>>>>>>>>
>>>>>>>> However, in spite of the differences, they do share a common 
>>>>>>>> fieature:
>>>>>>>> your HHH is not a part of any of the proofs.
>>>>>>>
>>>>>>> All of the conventional proofs of the HP assume that
>>>>>>> there is an *input D* that can actually do the opposite
>>>>>>> of whatever value that HHH returns.
>>>>>>
>>>>>> Depends on what you mean by "conventional". If you merely mean proofs
>>>>>> that apply ordinary logic then there are proofs with a different
>>>>>> strategy. If you mean only proofs that use the same strategy that
>>>>>> Turing used then you are closer to the truth. But there is no 
>>>>>> assumption
>>>>>> about the exstence of such D. Its existence is proven.
>>>>>
>>>>> In seems that way until you pay much closer attention.
>>>>
>>>> No, it seems that way when you pay enough attention.
>>>>
>>>>> int main()
>>>>> {
>>>>>    DDD(); // The HHH that DDD calls cannot report on the
>>>>> }        // behavior of its caller because it cannot see
>>>>>           // is caller.
>>>>
>>>> If HHH is correctly constructed it does see DDD and everything DDD
>>>> calls. Nothing else is relevant.
>>>>
>>>
>>> Try to show how HHH can see anything about its own caller
>>> when HHH is not even allowed to look at its caller.
>>>
>>
>> It is irrelevant whether DDD is the caller of HHH or not.
>> int main()
>>   {
>>      HHH(DDD);
>>      return;
>>   }
>>
>> Now HHH is not called from HHH, but has the same input and it should 
>> see that DDD includes the Halt7.c code, which aborts, so it should 
>> see: a halting program.
> 
> In other words you fail to understand that
> halting requires reaching a final halt state.
> 

But DDD does reach a final state when it is run. Note, each program is 
run in its own environement, so the fact that we don't have that 
environment here isn't an issue.

Of course, if you want it in both, just make main be:

int main()
   {
     HHH(DDD);
     DDD();
     return;
   }

and that shows that HHH(DDD) will return 0, but DDD() will halt, and 
thus HHH was just wrong.

Sorry, you are just proving your ignorance of what the words actualy 
mean, becuase you have beleived your own lies.