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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: How do computations actually work?
Date: Mon, 26 May 2025 20:02:02 +0200
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Op 26.mei.2025 om 18:26 schreef olcott:
> 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.
> 

In other words you fail to understand that if HHH fails to reach the 
specified final state, that is an bug in HHH.