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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: How do computations actually work?
Date: Wed, 28 May 2025 10:38:54 +0300
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On 2025-05-27 15:41:23 +0000, olcott said:

> On 5/27/2025 3:31 AM, Mikko wrote:
>> On 2025-05-26 16:26:47 +0000, olcott said:
>> 
>>> 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.
>> 
>> It is a sin to present false claims about tother peoples understanding.
> 
> You have acted like you don't know this.

You always act like a liar. Not like a good liar but like a stupid liar.

-- 
Mikko