Re: We finally know exactly how H1(D,D) derives a different result than H(D,D)

On 3/9/2024 3:07 AM, Mikko wrote:

On 2024-03-08 16:09:58 +0000, olcott said:
 

On 3/8/2024 9:29 AM, Mikko wrote:

On 2024-03-08 05:23:34 +0000, Yaxley Peaks said:
>

With all of these extra frills, aren't you working outside the premise
of the halting problem? Like how Andre pointed out.

>
Yes, he is.
>

The halting problem concerns itself with turing machines and what you
propose is not a turing machine.

>
That is true. However, we can formulate similar problems and proofs
for other classes of machines.
>

>
I am working on the computability of the halting problem
(the exact same TMD / input pairs) by a slightly augmented
notion of Turing machines as elaborated below:
>
Olcott machines are entirely comprised of a UTM + TMD and one
extra step that any UTM could perform, append the TMD to the
end of its own tape.

 An important question to answer is whether a Turing machine can
simulate your machines.

Olcott machines are entirely comprised of a UTM + TMD and one
extra step that any UTM could perform, append the TMD to the end
of its own tape.
Yes but this machine itself cannot be an Olcott machine. It may or may
not make a difference that this machine cannot be an Olcott machine.

 Another interesting question is whether your machines can solve
their own halting problem.
 

I don't know the details of this yet I do know that the Linz Ĥ
can only fool itself and not any external H. It is also the case
that the Linz machine must either halt or fail to halt and in
either case H ⟨Ĥ⟩ ⟨Ĥ⟩ can see this.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer