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

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.

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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.

 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.

 That someone can intentionally break an otherwise correct
halt decider

It always gives exactly the same answer as the working one, so how is it possibly broken?
provides zero evidence that such a decider does

not exist.
 Any changes made to Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ would have the result that
Ĥ ⟨Ĥ⟩ halts or fails to halt and H ⟨Ĥ⟩ ⟨Ĥ⟩ can see that.