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From: Richard Heathfield <rjh@cpax.org.uk>
Newsgroups: comp.theory
Subject: Re: Turing Machine computable functions apply finite string
 transformations to inputs VERIFIED FACT
Date: Wed, 30 Apr 2025 19:11:28 +0100
Organization: Fix this later
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On 30/04/2025 18:28, olcott wrote:
> On 4/30/2025 10:46 AM, Richard Heathfield wrote:
>> On 30/04/2025 16:15, olcott wrote:
>>> On 4/29/2025 5:03 PM, Richard Heathfield wrote:
>>>> On 29/04/2025 22:38, olcott wrote:
>>>>
>>>> <snip>
>>>>
>>>>>
>>>>> int DD()
>>>>> {
>>>>>    int Halt_Status = HHH(DD);
>>>>>    if (Halt_Status)
>>>>>      HERE: goto HERE;
>>>>>    return Halt_Status;
>>>>> }
>>>>>
>>>>> HHH is correct DD as non-halting BECAUSE THAT IS
>>>>> WHAT THE INPUT TO HHH(DD) SPECIFIES.
>>>>
>>>> You're going round the same loop again.
>>>>
>>>> Either your HHH() is a universal termination analyser or it 
>>>> isn't. 
>>>
>>> The domain of HHH is DD.
>>
>> Then it is attacking not the Halting Problem but the Olcott 
>> Problem, which is of interest to nobody but you.
>>
> 
> Because you don't pay any attention at all
> you did not bother to notice that I have never been
> attacking the Halting Problem only the conventional
> Halting Problem proof.

HHH(DD) doesn't attack the Halting Problem proof any more than a 
bucketful of marshmallows attacks Fort Knox. It fails to model 
the problem correctly.

> THE IMPOSSIBLE INPUT IS REJECTED AS NON-HALTING.

If HHH rejects DD, it fails to meet the spec for a decider, which 
doesn't have licence to reject programs.

If it reports on DD as halting, DD fails to meet the spec for an 
input that you appear to agree is impossible to report on.

-- 
Richard Heathfield
Email: rjh at cpax dot org dot uk
"Usenet is a strange place" - dmr 29 July 1999
Sig line 4 vacant - apply within