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

On 3/7/24 1:05 PM, olcott wrote:

H1(D,D) maps its input + its own machine address 00001422 to its output.
  H(D,D) maps its input + its own machine address 00001522 to its output.
Thus both H1 and H are computable functions of their input.

And thus you are admitting that Neither H or H1 are actually correct Halt Deciders, as Halt Deciders must be only a function of the description of the Compuation to be decided.

 Turing machines don't even have the idea of their own machine
address so this exact same thing cannot be Turing computable.

And it isn't a Halt Decider even in Olcott machines as the algorithm is shown to vary by a parameter that it isn't allowed to vary to be a Halt Decider.

 Olcott machines entirely anchored in Turing machine notions
can compute the equivalent of H1(D,D) and H(D,D).
 Because Olcott machines are essentially nothing more than
conventional UTM's combined with Conventional Turing machine
descriptions their essence is already fully understood.
 The input to Olcott machines can simply be the conventional
space delimited Turing Machine input followed by four spaces.
 This is followed by the machine description of the machine
that the UTM is simulating followed by four more spaces.
 When this input is ignored Olcott machines compute the
exact same set as Turing machines.
 Unlike Turing machines, Olcott machines have the basis to
determine that they have been called with copies of their
own TMD.
 Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqy ∞ // Ĥ applied to ⟨Ĥ⟩ halts
Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.Hq0 ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn   // Ĥ applied to ⟨Ĥ⟩ does not halt
 With Olcott machines Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> and H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> do
not have the same inputs thus can compute different outputs
when they do not ignore their own TMD.
 

THen you build H^ wrong. Of course with your change in mechanics, the H^ that needs to be generated will be a bit different.
H^ needs to have a copy of the algorithm of the decider at that point, and give it exactly the same input as the decider will get.
Since H will get on its tape the string (H^) (H^) (H), that is exactly what H^ (H^) must put on its tape, so since H^ starts its tape with the values (M) (H^), it must duplicate the (M), and replace the (H^) with the description of the version of H it is going to confound. (even though that isn't the description of the copy at H^.H).
Note, while your master UTM may make sure that no Top level machine is given the wrong description, there is nothing stopping a machine from faking the data to a submachine of it.
You are just not thinking the problem out far enough.
You are apparently not smart enough to figure out what this change needed to be to handle you new definition of Olcott machines, but I was able to do that in just a few minutes of thinking.