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

On 3/7/2024 5:18 PM, Richard Damon wrote:

On 3/7/24 3:02 PM, olcott wrote:

On 3/7/2024 4:32 PM, Richard Damon wrote:

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.

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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.
>

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It turns out that both H(D,D) and H1(D,D) do correctly determine
whether or not they must abort the simulation of their input.

 Which isn't the halting question, so you are LYING.

As I completely explained yet you persistently ignore the
halting question can only be correctly answered indirectly
otherwise inputs that contradict the decider that is embedded
within these inputs have no answer at all.

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That you or others consider this somehow improper does not change
the verified fact that they both correctly determine whether or
not they must abort their simulation.

 Which isn't the Halting Question, which you claim you are working on, so you are just LYING.
 

Already fully explained many many times (including above)
yet your ignorance is very persistent.

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It is also the case that both H1(D,D) and H(D,D) are a pure function
of their inputs when we construe their own machine address to be an
element of these inputs.

 Which means they are not computing the Halting Function, which isn't a function of the decider, so again, you are LYING.
 

Already fully explained many many times (including above)
yet your ignorance is very persistent.

>

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

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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.
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Olcott machines entirely anchored in Turing machine notions
can compute the equivalent of H1(D,D) and H(D,D).
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Because Olcott machines are essentially nothing more than
conventional UTM's combined with Conventional Turing machine
descriptions their essence is already fully understood.
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The input to Olcott machines can simply be the conventional
space delimited Turing Machine input followed by four spaces.
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This is followed by the machine description of the machine
that the UTM is simulating followed by four more spaces.
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When this input is ignored Olcott machines compute the
exact same set as Turing machines.
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Unlike Turing machines, Olcott machines have the basis to
determine that they have been called with copies of their
own TMD.
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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
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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.
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>

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THen you build H^ wrong. Of course with your change in mechanics, the H^ that needs to be generated will be a bit different.
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That Olcott machines always know their own TMD is unconventional.

 And breaks much of the background of Turing Machines,

Not at all. Not in the least little bit.
Olcott machines are 100% fully specified
in terms of Turing machines.

so if you what to use ANY establish property of Turing Machine, you must include that now extra data EXPLICITLY.
 

It is already explicitly included in the definition of an Olcott machine.

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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.

 But not the RIGHT computation.
 

Not the same computation that conventional wisdom expects.

Right answer to the wrong question is still a WRONG ANSWER.
 

The right answer to a different question that can always
be construed as the right answer to the original question
by H ⟨Ĥ⟩ ⟨Ĥ⟩ is the right answer.

So, you are just shown again to be a LIAR.
 

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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.
>

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Both Ĥ and H have a copy of their own non-identical TMD's.

 Yep, and H^ can give its H^.H the same TMD of the machine that will decide it to its copy of H's algorithm,

You are not saying that Ĥ does not have its own TMD
you are only saying that it can screw this up too.
No matter what Ĥ does it will either halt or fail to
halt in a way that can be determined by the actual H.

so it gets the same answer as that H wil produce, and thus that H will give the WRONG answer to the Halting Question,
 

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Since H will get on its tape the string (H^) (H^) (H), that is exactly what H^ (H^) must put on its tape,

It will not get this, where the Hell did you get this from?
Ĥ ⟨Ĥ⟩ gets only two inputs and both of them are ⟨Ĥ⟩.
The latter one is appended to its tape by the master UTM.
Ĥ never gets any separate <H>.

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Not at all. Olcott machines are exactly the same as Turing machines
yet always have a copy of their own machine description appended to
the end of their own tape. That you assert that H must have a copy of
some other TMD at the end of its tape is quite nutty.

 WHAT WAS WRONG WITH WHAT I SAID.
 Where did I say that a machine when started had anything else.
 The SUB-MACHINE H^.H can have on its tape whatever H^ wants to put there, if that puts in the "Your TMD" slot the description of some other verison of it, that is fine and you can't stop it.
 

Ĥ ⟨Ĥ⟩ cannot possibly put a separate <H> anywhere on its tape
it has no access to any such separate <H>.

So H^.H (H^) (H^) (H) will give exactly the same answer as the top level
H (H^) (H^) (H) will give.  And thus H^ will make H give the wrong answer for the Halting Problem.
 

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so since H^ starts its tape with the values (M) (H^),

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Not at all. The only Ĥ that I am talking about only has
its own TMD for its input.

 H^ IN GENERAL, can have the description of ANY machine on its tape.
 The one case we look at, M = H^, which is its own.
 so H^ (H^) (H^)  {last one added by the Master UTM, or H}
goes to H^.Hq0 (H^) (H^) (H) by duplicating its input data, removing its own description and adding the description of the H it is tragetig.
 

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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).
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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.
>

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No matter how Ĥ screws things up it will either halt or fail
to halt. No matter how much Ĥ screws itself up none of these
things have any effect what-so-ever on H.

 But what it does is ALWAYS the opposite of what H will say.

Not all all. not even once. Ĥ only ever does the opposite of whatever
its own internal Ĥ.H says. it has no access to any external <H>.

 H^ gets to know this, as it has a copy of that H, and it knows what the tape to H will look like when it is asked about itself.

It has no idea that it has a copy of any external <H>.

 You can't (successfully) argue with facts.

Not to someone that doesn't care about facts and only
cares about rebuttal.

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You are just not thinking the problem out far enough.
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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.

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I thought this through more deeply than you have and you cannot
show otherwise by providing correct reasoning why my ideas will
not work.
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*Like most people you conflate unconventional with impossible*
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 So, why doesn't H^ get the same answer as H will give when it gives its exact copy and exact copy of the input that H will get?
 

*Like I said and you have not fully integrated into your thinking*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> does indeed map to a different output than H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>.
I do still really appreciate all of your feedback.
You do show a very good understanding of these things.

YOU are the one that assumes the impossible, so your logic trues out to be just the mystical product of Fairy dust powered Unicorns, and not anything that actually works.
  Your FAILURE to answer the questions just proves your guilt.

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