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On 3/9/2024 3:49 PM, immibis wrote:If two machines do not behave the same on the same inputOn 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: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.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*Specifications, not actual behavior until the existance of such an H is shown.
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
IF taken as actual behavior, then it is conditional on such an H existing.
Execution trace of Ĥ applied to ⟨Ĥ⟩It NEEDS to in order to meet its specification
(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 DOESN'T unless its algorithm says it does,Yes. (Notice that I am agreeing with you, yet never do that with me)
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)
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*
If it halts then Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ transitioned to Ĥ.Hqn.
must have the same behavior on the same input.
This may not apply when these machines having identicalMeanings of the words do apply.
states and identical inputs:
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