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

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

>
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.

>
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.
>
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
>
When this is executed in an Olcott machine then
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <Ĥ> is a different computation than H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>

 WHY?
 The Master UTM will not change the usage at H^.H, because that is just an internal state of the Machine H^
 

The ONLY thing that the master UTM does differently is append
the TMD to the TMD's own tape.

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

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

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

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.
>

 But that isn't what H^.H gets called with.

Olcott machines always append the TMD to the end of the tape
of this TMD.
This means that they are an actual input to every TMD that
does not ignore them.

 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.
 

Ĥ cannot possibly do this because it has no access to nor
even knows the existence of any external ⟨H⟩.

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

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