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

On 3/7/2024 8:06 PM, André G. Isaak wrote:

On 2024-03-07 16:02, olcott wrote:
 

That Olcott machines always know their own TMD is unconventional.
>
That their own TMD is correctly construed as an additional input
to their computation (whenever they don't ignore it) does provide
the reason why Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> can compute different
results and still be computations.

 It's also the reason why you approach is fundamentally flawed. Putting aside the question of whether your proposal is workable (or even sane), if your 'Olcott Machines' automatically supply the machines they emulate with a copy of their own machine descriptions, then you are no longer working on the halting problem.
 

*Thanks for your feedback, I really appreciate it*
All actual feedback helps me to debug my words.
I would phrase that more precisely that I am no longer working
on the computability of the halting problem by Turing 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.
Olcott machines that ignore this extra input compute the exact
same set of functions that Turing machines compute.

The halting problem asks, is it possible to construct a TM X that, given a description of a second TM Y and an input string Z *and* *only* *that* *input* *to* *work* *with*, is it possible for X to determine whether Y applied to Z halts.

Sure. When an Olcott machine correctly decides some of the elements
of the TMD/Input pairs that a Turing machine cannot decide then Church-
Turing would refuted.

Asking whether it is possible to construct a TM X which, given a description of a second TM Y, and input string Z, *and* a description X, can X determine whether Y applied to Z halts, is an *entirely* different question.
 

Sure. None-the-less when an Olcott machine computes everything that
a Turing Machine can compute (by definition) and can compute TMD/Input
pairs that a TM cannot compute then Church-Turing would refuted.

The answer to these two questions may well be entirely different, and the answer to the second question tells us absolutely nothing about the answer to the first, which is the only thing the halting problem is concerned with.
 André
 

--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer