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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: How do computations actually work?
Date: Thu, 29 May 2025 07:05:47 -0400
Organization: i2pn2 (i2pn.org)
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On 5/28/25 3:55 PM, olcott wrote:
> On 5/28/2025 2:13 PM, Fred. Zwarts wrote:
>> Op 28.mei.2025 om 16:31 schreef olcott:
>>> On 5/28/2025 2:36 AM, Mikko wrote:
>>>> On 2025-05-27 15:40:33 +0000, olcott said:
>>>>
>>>>> On 5/27/2025 3:29 AM, Mikko wrote:
>>>>>> On 2025-05-26 16:40:25 +0000, olcott said:
>>>>>>
>>>>>>> On 5/25/2025 10:46 AM, Fred. Zwarts wrote:
>>>>>>>> Op 25.mei.2025 om 16:50 schreef olcott:
>>>>>>>>> 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.
>>>>>>>>>
>>>>>>>>> int main()
>>>>>>>>> {
>>>>>>>>>    DDD(); // The HHH that DDD calls cannot report on the
>>>>>>>>> }        // behavior of its caller because it cannot see
>>>>>>>>>           // is caller.
>>>>>>>>>
>>>>>>>>> Even if HHH could see and report on the behavior of
>>>>>>>>> its caller because its caller is not its input this
>>>>>>>>> too is no good.
>>>>>>>>
>>>>>>>> It seems that way to you, until you pay somewhat closer attention.
>>>>>>>
>>>>>>> The HHH(DDD) must report on the behavior that its actual input
>>>>>>> actually specified CANNOT BE VIOLATED.
>>>>>>
>>>>>> Of course it can. In fact HHH does violate that. DDD specifies a 
>>>>>> halting
>>>>>> behaviour but HHH reports that DDD specifies a non-halting behaviour.
>>>>>> That is a violation of that rquirement.
>>>>>
>>>>> If DDD simulated by HHH stops running for any
>>>>> reason besides reaching its own "ret" instruction
>>>>> final halt state THEN DDD HAS NOT HALTED.
>>>>
>>>> Irrelevant. The requirement is that a halt decider predicts whether the
>>>> complete execution of the computation described by the input will halt.
>>>>
>>>
>>> Halting is defined as reaching a final state and
>>> terminating normally.
>>
>> So, according to this definition, every simulator that aborts after 
>> one instruction is correct to report a non-halting program.
>>
>>>
>>> int main()
>>> {
>>>    DDD(); // The HHH that DDD calls cannot
>>> }        // see the behavior of its caller
>>>
>>> *That is incorrect*
>>> A termination analyzer must report on the basis
>>> of the behavior that its input specifies and does
>>> not give a rat's ass about the behavior of its caller.
>>
>> The caller is not of any interest. The analyser must report on the 
>> input. In this case the anaylzer is given a pointer to memory as 
>> input. This contains the code of DDD. That code has addresses, among 
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