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On 3/9/2024 3:49 PM, immibis wrote:Nope, the machine execute INDEPENDENTLY of each other,On 9/03/24 22:30, olcott wrote:I generally agree that a pair of identical machinesOn 3/9/2024 3:17 PM, Richard Damon wrote:>On 3/9/24 10:33 AM, olcott wrote:Yes. (Notice that I am agreeing with you, yet never do that with me)*Verified fact that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ and H ⟨Ĥ⟩ ⟨Ĥ⟩ have different behavior*>
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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
Specifications, not actual behavior until the existance of such an H is shown.
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IF taken as actual behavior, then it is conditional on such an H existing.
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Execution trace of Ĥ applied to ⟨Ĥ⟩
(a) Ĥ.q0 The input ⟨Ĥ⟩ is copied then transitions to Ĥ.H
(b) Ĥ.H applied ⟨Ĥ⟩ ⟨Ĥ⟩ (input and copy) simulates ⟨Ĥ⟩ applied to ⟨Ĥ⟩
(c) which begins at its own simulated ⟨Ĥ.q0⟩ to repeat the process
*This proves that Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ must abort its simulation*
It NEEDS to in order to meet its specification
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>It DOESN'T unless its algorithm says it does,Yes. (Notice that I am agreeing with you, yet never do that with me)
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>If it just fails to answer, then it has failed to be a correct Halt Decider.Yes. (Notice that I am agreeing with you, yet never do that with me)
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>The fact that you reach this conflict in actions, is the reason Halt Deciding is uncomputable.>
*No. We know that Ĥ ⟨Ĥ⟩ either halts or fails to halt*
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If it halts then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn.
If Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn then H ⟨Ĥ⟩ transitioned to H.qn or else Ĥ is the wrong Ĥ or you can't read instructions.
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must have the same behavior on the same input.
This may not apply when these machines having identical
states and identical inputs:
(a) Are out-of-sync by a whole execution trace or
(b) When one of the machines is embedded within another machineNope.
that would cause this embedded machine to have recursive
simulation that the non-embedded machine cannot possibly have.
*I think that the actual difference is the latter case because*Why not? H -> H^ ... H^.H -> H^ ... H^.H is getting stuck in a recusive simulation.
*we have the exact same issue when the infinite loop is removed*
Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
*The above two behaviors are different for identical*Which just shows you don't understand what you are saying.
*machines with identical inputs*
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