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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: Tue, 29 Apr 2025 23:20:09 +0100
Organization: Fix this later
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On 29/04/2025 22:44, olcott wrote:
> On 4/29/2025 3:23 PM, Richard Heathfield wrote:
>> On 29/04/2025 20:57, olcott wrote:
>>> On 4/29/2025 10:33 AM, Fred. Zwarts wrote:
>>
>> <snip>
>>
>>>> makes it impossible for H to see the behaviour of P(D).
>>>> The behaviour of P(D) does not change, but H does not see it.
>>>
>>> H MUST REPORT ON THE BEHAVIOR THAT IT DOES SEE
>>
>> H has the whole P tape and the whole D tape at its disposal. 
>> There is nothing it can't inspect.
>>
> 
> int DD()
> {
>    int Halt_Status = HHH(DD);
>    if (Halt_Status)
>      HERE: goto HERE;
>    return Halt_Status;
> }
> 
> When P has a pathological relationship
> to H, the input to H(P,D) DOES NOT HALT.

Presumably you mean that the Turing machine on tape P, given tape 
D as input, doesn't halt.

But if that's true, then we can construct a pathological meta-P, 
so to speak, and arrange it so that it /does/ halt. When we feed 
/that/ to your HHH, it will now give the wrong answer.

>> Not being able to see how P behaves in simulation is no excuse 
>> for getting the answer wrong. 
> 
> THEN...
> I am thinking of the sum of two integers
> not telling you what these integers are
> is no excuse for you not providing their correct sum.

Absolutely.

 From my perspective, with the information I have available, I 
must fail to complete the task, because the answer is incomputable.

Similarly, we cannot write an H that, given arbitrary P and D, 
always correctly reports on whether feeding D to P would result 
in P halting. It can't be done. It's incomputable. That's the 
whole point!

The parallel is not /quite/ perfect, though, because H can see 
the tapes (P and D), and can examine them in their entirety.

Unfortunately, this isn't enough. We still can't guarantee to 
deduce whether P(D) would halt.

>> If because of limitations in H it fails to spot behaviour that 
>> would have changed its report on P(D), that just means that H 
>> is broken.
>>
>> But don't bother trying to fix it. Turing has already proved 
>> that you can't.
>>
> 
> You are simply not bothering to pay complete attention.

I'm ignoring most of your nonsense, if that's what you mean.

What you're ignoring is mathematical rigour.



-- 
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