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

On 3/8/2024 1:50 AM, Richard Damon wrote:

On 3/7/24 11:21 PM, olcott wrote:

On 3/8/2024 12:49 AM, Richard Damon wrote:

On 3/7/24 10:09 PM, olcott wrote:

On 3/7/2024 11:36 PM, Richard Damon wrote:

On 3/7/24 9:18 PM, olcott wrote:

Olcott machines can compute the difference between H1(D,D) and H(D,D)
because Olcott machines can correctly determine whether or not they
themselves were called in recursive simulation.
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For H1(D,D) and H(D,D)it is the extra inputs of 00001422 and 00001522
For H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> the extra inputs are shown.
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When I keep repeating the above a few hundred more times
you will quit immediately forgetting that I ever said it.

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All you have done is show why your H / H1 have been disqualified from being Halt Deciders.

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Not at all. Not in the least little bit.
I have proved that a slight reconfiguration of Turing machines
defines a machine that is exactly a Turing machine except can
always correctly determine whether or not itself is called in
recursive simulation.

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Nope.
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You tried, but failed.
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All that you have done is disagree with verified facts.
You always said that H1(D,D) and H(D,D) cannot compute different outputs. I proved exactly how they do this and how it is computable
and you simply ignore what I say.

 They can't if they are copies of the same computation.

Exact copies of the same machine with different inputs can and
do compute different results. I really should not have to bring
this up again and again all day long for you to notice that
I said it at least once.

 You are just proving that they aren't, as either they use a "Hidden Input" in your x86UTM claiming to be Turing Equivalent model, or they now have a different input in the Olcott-Machine model.
 

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When H1 and H both have access to their own machine address this is equivalent to H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> and Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ>

 Right, so they are getting different inputs, so they can create different outputs,

Yes.

but that also means they are not Halt Deciders, as they are USING that extra data that Halt Deciders can't use (at least an be a CORRECT Halt Decider).

They are using extra inputs that Turing machines cannot use.
Thus my proof does not show that halting is Turing computable.
Instead my proof shows that Church-Turing is not true.
Olcott machines compute everything that Turing Machines compute
and have the extra ability to correctly detect that they themselves
are about to simulate themselves in recursive simulation.

 You are just showing how little you understand of the field and the meaning of its words.
 

You have not correctly pointed out even a single mistake.

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Your H and H1 likely will never answer when given the properly designed H^ input.
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You are just a total failure, again.
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And are just lying about it because you refuse to face the facts.
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They are not Computations using only the description of the Computation to be decided on.
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This PROVES that they are not correct Halt Deciders.
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And you are proved to be just a pathetic and pathological liar.

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