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On 3/9/24 2:15 PM, olcott wrote:It is objectively true that Ĥ.H can get stuck in recursiveOn 3/9/2024 3:49 PM, immibis wrote:Nope, the machine execute INDEPENDENTLY of each other,On 9/03/24 22:30, olcott wrote:>On 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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I generally agree that a pair of identical machines
must have the same behavior on the same input.
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This may not apply when these machines having identical
states and identical inputs:
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(a) Are out-of-sync by a whole execution trace or
The only thing that shows this "sync" issue is that the simulation of machines, where one machine simulating another might be at different stages.
You keep on confusing the simulation of the machine with the actual behavior of the
>Nope.
(b) When one of the machines is embedded within another machine
that would cause this embedded machine to have recursive
simulation that the non-embedded machine cannot possibly have.
Not posssible.
Please try to show how this can happen.
ACTUAL INSTRUCTION that differ in path.
You seem to NEED for this to be true, but it isn't
>Why not? H -> H^ ... H^.H -> H^ ... H^.H is getting stuck in a recusive simulation.
*I think that the actual difference is the latter case because*
*we have the exact same issue when the infinite loop is removed*
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Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ can possibly get stuck in recursive simulation and
H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly get stuck in recursive simulation.
Just like if F1 calls f2, and f2 calls f2, then f1 will get stuck in a "infinite recursive call loop" even though f1 isn't part of the loop>Which just shows you don't understand what you are saying.
*The above two behaviors are different for identical*
*machines with identical inputs*
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