Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--

On 3/7/2024 9:53 PM, Richard Damon wrote:

On 3/7/24 7:29 PM, olcott wrote:

On 3/7/2024 8:34 PM, Richard Damon wrote:

On 3/7/24 6:02 PM, olcott wrote:

On 3/7/2024 7:35 PM, immibis wrote:

On 7/03/24 18:05, olcott wrote:

On 3/7/2024 8:47 AM, immibis wrote:

On 7/03/24 03:40, olcott wrote:

On 3/6/2024 8:22 PM, immibis wrote:

On 7/03/24 01:12, olcott wrote:

On 3/6/2024 5:59 PM, immibis wrote:

On 7/03/24 00:55, olcott wrote:

Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.Hqn
Correctly reports that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation.
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H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* H.qy
Correctly reports that H ⟨Ĥ⟩ ⟨Ĥ⟩ need not abort its simulation.

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What are the exact steps which the exact same program with the exact same input uses to get two different results?
I saw x86utm. In x86utm there is a mistake because Ĥ.H is not defined to do exactly the same steps as H, which means you failed to do the Linz procedure.

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Both H(D,D) and H1(D,D) answer the exact same question:
Can I continue to simulate my input without ever aborting it?
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Both H(D,D) and H1(D,D) are computer programs (or Turing machines). They execute instructions (or transitions) in sequence, determined by their programming and their input.

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Yet because they both know their own machine address
they can both correctly determine whether or not they
themselves are called in recursive simulation.

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They cannot do anything except for exactly what they are programmed to do.

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H1(D,D) and H(D,D) are programmed to do this.
Because H1(D,D) simulates D(D) that calls H(D,D) that
aborts its simulation of D(D). H1 can see that its
own simulated D(D) returns from its call to H(D,D).
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An Olcott machine can perform an equivalent operation.
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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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To make the Linz proof work properly with Olcott machines, Ĥ should search for 4 spaces, delete its own machine description, and then insert the description of the original H. Then the Linz proof works for Olcott machines.

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That someone can intentionally break an otherwise correct
halt decider

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It always gives exactly the same answer as the working one, so how is it possibly broken?
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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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When this is executed in an Olcott machine then
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> is a different computation than H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>

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WHY?
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The Master UTM will not change the usage at H^.H, because that is just an internal state of the Machine H^
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The ONLY thing that the master UTM does differently is append
the TMD to the TMD's own tape.

 Right, so it gives H and H^ that input.
 H^ can erase that input and replace it.
 

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At that point we have the IDENTICAL set of transitions (with just an equivalence mapping of state numbers) as H will have, and the EXACT same input as H

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it is stipulated by the definition of Olcott machines
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> // last element is <Ĥ> (not H)

 Nope.
 H^.H isn't a "Olcott Machine" it is a sub-machine of H^

You already know that there is no such thing as sub-machines of
Turing machines. Turing machines have a set of states and a tape.
They have no sub-machines.

 Your Master UTM can't touch the tape here or it totally breaks the system.
 

The master UTM is merely an ordinary UTM that always appends
the slaves TMD to the slaves tape. That is all that there is
to the master UTM.

It can only add the description at the beginning, as it doesn't know what state in that machine are "sub-machines" and what are just parts of the machine running.
 

An ordinary UTM writes the input to the slaves tape.
The master UTM of Olcott machines simply adds one more step to this.
The master UTM is merely an ordinary UTM that always appends
the slaves TMD to the slaves tape. That is all that there is
to the master UTM.

You don't seem to understand much about Turing machines.
 I seem to remember you bailed out after like the first part of the first lesson.
 

You can't point out an errors because there are no errors.
You keep saying things at least slightly incorrectly like
sub-machines of Turing machines, there is no such thing.

   H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> // last element is <H> (not Ĥ)
That the last element of the input to Ĥ.H and H <is> different.
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 Until the code in H^ between q0 and H^.H changes it.

I have no idea what you are saying yet the only category that
it can be is that Ĥ tries to thwart its own internal termination
analyzer.

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No matter how Ĥ ⟨Ĥ⟩ screws itself up this can have no
effect on H ⟨Ĥ⟩ ⟨Ĥ⟩.
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The Olcott machine H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> can see that it does not
call itself in recursive simulation.

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As will the machine at H^.H, since the description it thinks it is doesn't match the machine that gets called.

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A simply string comparison of the finite strings
⟨Ĥ⟩ and <Ĥ> proves that they are the same.
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The Olcott machine Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> can see that it calls
itself in recursive simulation unless it discards this
ability. In either case it cannot fool H, it either halts
or fails to halt.
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But that isn't what H^.H gets called with.

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Olcott machines always append the TMD to the end of the tape
of this TMD.

So H^ STARTS with <H^> at the end. The Tape, BY DEFINITION, is writable.
 That is sort of one problem with your x86UTM, its "tape" equivalent was Read Only since it was executable code space.
 

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This means that they are an actual input to every TMD that
does not ignore them.

 Right, and H^ can change it.
 

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The code between H^.q0 and H^.H removes the <H^> and replaces it with <H>, so the copy of H at H^.H can't tell the difference.
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Ĥ cannot possibly do this because it has no access to nor
even knows the existence of any external ⟨H⟩.

 Of course it know about the ONE H that it is on a mission to make wrong.

Most of the halting problem proofs fail to understand
that an actual Turing machine cannot actually call
another external Turing machine. This means that it
cannot actually confound an external Turing machine.

 So that H, and only that H gets confonded.
 You sure seem dumb about that.
 

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You don't seem to understand what a Turing Machine CAN do.
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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