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

On 3/8/2024 11:47 AM, Richard Damon wrote:

On 3/8/24 9:33 AM, olcott wrote:

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.

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

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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.

 Yes, but they are given the same input, at least when yo follow the rules an not use a strawwman.
 The DEFINITION of H^ is that it uses an exact copy of the algorithm it is refuting, and gives it the exact same input that it would have when deciding on it.
 The fact that you just rotely parrot the H^ for Turing Mchines when you try to apply it to Olcott-Machine just says you don't understand Linz's proof.
 

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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>

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

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Yes.

 So H1 doesn't show that the algorithm of H gave the right answer.
 

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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).

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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.

 Not "Can't", but weren't given.

Olcott machines make providing this input mandatory thus
closing the loophole that Turing machines leave open.

 You are just proving yourself to be an idiot.

My Mensa test showed I am in the top 3%.

Remember, the defition of a Computation includes ALL the inputs given, so your Olcott machines are not doing the same computation as the Turing Machines, and the Turing Machines COULD do that same computation if given the extra data.

That is correct except:
Olcott machines make providing this input mandatory thus
closing the loophole that Turing machines leave open.

All you have done is LIMIT what Olcott machines can be shown to compute, and to claim they do the same computation as the Turing Machine they appear to be, you have to first proe they don't use that extra information.
 

Olcott machines (by definition) compute all of the functions
that Turing machines compute and have the extra ability to
always detect when they are about to simulate themselves in
recursive simulation.

Your whole problem is you don't understand what a "Compuation" is.

I consistently proved otherwise and you have not correctly
pointed out a single error.
Even H(D,D) really does prove that from its own POV its input
specifies a non-halting computation. So I never made a mistake
on that either. What I said about that is merely unconventional
it is not incorrect.

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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.

 Nope.

I have proven this and you provided no evidence to the contrary.
The above reply is exactly what I mean by you using dogma instead
of reasoning.
None-the-less when you do pay attention to a key detail
this does provide a good basis for me to debug my words.
It is not that my words were incorrect, they were simply
too difficult to understand.

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You are just showing how little you understand of the field and the meaning of its words.
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You have not correctly pointed out even a single mistake.

 Sure I have, you are just to stupid to understand, because you don't know what you are talking about.

All of the "mistakes" that you pointed out have been false
assumptions that are based on your lack of attention to
the details of what I am saying.

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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