On 3/8/2024 3:57 PM, Richard Damon wrote:
On 3/8/24 1:34 PM, olcott wrote:
On 3/8/2024 2:29 PM, Richard Damon wrote:
On 3/8/24 11:17 AM, olcott wrote:
On 3/8/2024 12:40 PM, Richard Damon wrote:
On 3/8/24 10:11 AM, olcott wrote:
On 3/8/2024 12:00 PM, Richard Damon wrote:
On 3/8/24 7:59 AM, olcott wrote:
On 3/8/2024 5:26 AM, Mikko wrote:
On 2024-03-07 19:49:49 +0000, Dan Cross said:
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What is it? The olcott machine is a device that never halts and
generates infinite amounts of nonsense. As a perpetual motion
device with no discernable input and unbounded output, it is
believed that it violates the laws of thermodynamics.
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The olcott machine uses a hidden input.
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It is not hidden. The master UTM of Olcott machines simply
appends the TMD to the end of the simulated TMD's tape.
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Only those machines that need to see if themselves are
called in recursive simulation use this optional input.
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Which means they ADMIT they are doing a different computation then the Turing Machine they are derived from.
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So, there can not be an Olcott Machine that matches the signature of a Halt Decider.
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PERIOD
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And thus, you prove you have created another worthless field.
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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:
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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.
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Olcott machines that ignore this extra input compute the exact
same set of functions that Turing machines compute.
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Olcott machines can do something that no Turing machine can
possibly do correctly determine that they themselves are
called in recursive simulation.
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Nope.
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You have PROVED (by your definition of an Olcott Machine) that ANYTHING an Olcott machine can do, there exists a Turing Machine that does the same thing.
There is no conventional Turing machine that can possibly
know that it is about to simulate a copy of itself in
recursive simulation.
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It can know just as well as your Olcott machines, which apparently can only tell it the recusion is done by that EXACT same machine using the same description
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How it this?
Conventional Turing machines do not generally have access to their
own machine description and generally cannot even know that they
are being provided with their own machine description unless they
are Olcott machines where this is anchored in their fundamental
architecture.
But if the mapping includes the description, it can be given to it.
I am only going to respond to this one thread
(until I complete my design)
and will stop responding to this one too if you don't provide
very well thought out rebuttals or replies.
If a machine is given its own description unless the fundamental
architecture of the system guarantees that this is its true
description then the machine cannot rely on this being its own
description.
Anchoring it in the architecture means you now need to PROVE they don't use it if the mapping doesn't have it (like most don't)
Not at all as I just explained above.
So you have added to your work to prove the behavior of the machine.
I don't see how.
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So, H (H) (H) <H> (if all the H's use the same description can be detected), but not H (H^) (H^) <H> as the description of H at H^.H has different state numbering than H so the description will be different.
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Olcott machines only need to be able to detect that they themselves
about about to simulate a copy of themselves with their same input.
I am working out all of the details of this so I will be less responsive
to your many posts until I get this worked out.
So, the H / H^ pairing doesn't need to be able to do that?
Unless a machine can certainly know that it has its own machine
description as guaranteed by the architecture of this system it
cannot rely on this.
H needs to verify that it is not being called with its own
description. I am working out the details of what happens
when H ⟨H⟩ ⟨H⟩ <H> is invoked.
I thought that was the whole reason for adding it.
You just did it wrong.
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It would be best that you carefully study my future posts so that
you don't keep rebutting the same things that I have already fully
addressed. I generally spent a lot of time on your posts carefully
studying the exact words that you said. This is not very fruitful
when you do not do the same.
So, you are ok with the fact that H.q0 (H^) (H^) <H> will never return an answer?
*You are failing to stay within the boundaries of the stated scope*
The exact Linz Ĥ executed as an Olcott machine. Variations of the
exact Linz Ĥ will never be looked at until we have 100% complete
closure on the exact Linz Ĥ executed as an Olcott machine.
I cannot afford to tolerate the [change the subject] form of
rebuttal that wasted 15 years with Ben Bacarisse.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn // working out the details of how
therefore
H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> ⊢* H.qy // necessarily correct & details seem self-evident
Since it won't detect that H^ is H, and the H^.Hq0 (H^) (H^) <H> won't detect that it is about to simulate itself.
*You are failing to stay within the boundaries of the stated scope*
The exact Linz Ĥ executed as an Olcott machine.
I cannot afford to tolerate the [change the subject] form of
rebuttal that wasted 15 years with Ben Bacarisse.
I guess you are just admitting that you are just a liar that this solves the problem.
The fact that YOU aren't reading the replies seems evident, as YOU haven't given thought to the problem pointed out as you don't understand what the problem is, since you don't actually know how any of this works.
*You are failing to stay within the boundaries of the stated scope*
The exact Linz Ĥ executed as an Olcott machine.
I cannot afford to tolerate the [change the subject] form of
rebuttal that wasted 15 years with Ben Bacarisse.
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And since H^ can "lie" to that embedded H^.H about what its description is, that H can't tell that it is part of an H^ computation that is simulating an H^ computation.
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That subject must be postponed until after the Olcott refutation
of the exact Linz proof is either fully accepted by three people
or actual errors or gaps are found that cannot be addressed or
corrected.
So, how are you addressing the actual problems that I have pointed out.
*You are failing to stay within the boundaries of the stated scope*
The exact Linz Ĥ executed as an Olcott machine.
I cannot afford to tolerate the [change the subject] form of
rebuttal that wasted 15 years with Ben Bacarisse.
That by your current definition, neither H nor H^.H will abort their simulation of their input so we end up with an infinitely nested set of simulations.
*You are merely failing to pay attention to what I said*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn // working out the details of how
therefore
H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> ⊢* H.qy // necessarily correct & details seem self-evident
Just like how you kept ignoring that H^ could put <H> onto its tape.
*You are failing to stay within the boundaries of the stated scope*
The exact Linz Ĥ executed as an Olcott machine.
I cannot afford to tolerate the [change the subject] form of
rebuttal that wasted 15 years with Ben Bacarisse.
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Olcott machines make it impossible for a machine to not
know its own machine description.
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It know *A* version of its machine description, not ALL of them (as that is an infinte amount of data)
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The Olcott machine Linz H does correctly determine the halt status
of every Olcott machine Linz Ĥ the exactly matches the Linz Ĥ template.
Nope, not the Linz H^ templete that meets the SEMANTIC requirements of the template, not the syntax base on them being Turing Machines.
I cannot understand your words.
When the exact Linz H and the exact Linz Ĥ are executed as Olcott machines then the following is true.
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn // working out the details of how
therefore
H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> ⊢* H.qy // necessarily correct & details seem self-evident
When you say that it is not true because Turing Machines cannot do this
you are using the strawman deception to change the subject away from my
claim about Olcott machines.
-- Copyright 2024 Olcott "Talent hits a target no one else can hit; Geniushits a target no one else can see." Arthur Schopenhauer
Working out the details of the steps of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn
Re: Working out the details of the steps of Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> ⊢* Ĥ.Hqn