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On 3/9/2024 7:11 AM, immibis wrote:Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> always calculates the same answer that H ⟨Ĥ⟩ ⟨Ĥ⟩ <H> calculates.On 8/03/24 20:08, olcott wrote:Because Olcott machines are more powerful than Turing Machines they can correctly determine the halt status of this input: H ⟨Ĥ⟩ ⟨Ĥ⟩ <H>.On 3/8/2024 12:58 PM, immibis wrote:>On 8/03/24 19:12, olcott wrote:>>>
Unless it is an extra parameter it has no basis for doing this.
If it is an extra parameter then it is no longer the Linz proof.
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So it's impossible to make a Turing machine that writes 12345 onto its tape unless 12345 is a parameter?
That is a valid point.
It is impossible to get me to talk about that until
we first have full closure that the Linz H correctly
determines the halt status of the Linz Ĥ ⟨Ĥ⟩ when run
in the Olcott master UTM.
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When run in the Olcott master UTM as Olcott machines, the Linz proof does not work because it is designed for Turing machines. A different proof works.
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