On 8/1/2024 2:44 AM, Mikko wrote:Not "almost". Otherwise it is doing something different.
On 2024-07-31 17:27:33 +0000, olcott said:
On 7/31/2024 2:32 AM, Mikko wrote:
On 2024-07-30 14:16:20 +0000, olcott said:
On 7/30/2024 1:37 AM, Mikko wrote:
On 2024-07-29 16:16:13 +0000, olcott said:
On 7/28/2024 3:02 AM, Mikko wrote:
On 2024-07-27 14:08:10 +0000, olcott said:
On 7/27/2024 2:21 AM, Mikko wrote:
On 2024-07-26 14:08:11 +0000, olcott said:
Which dictionary (or other authority) disagrees?When we compute the mapping from the input to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
to the behavior specified by this input we know that embedded_H is >>>>>>> correct to transition to Ĥ.qn.
The meaning of "correct" in this context is that if the transition >>>>>> of embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ to Ĥ.qn is correct if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions
to H.qn but incorrect otherwise.
No you are wrong.
The common knowledge that a decider computes the mapping from its
input finite string...
This is almost always the same as the direct execution of the machine
represented by this finite string.
Dude. The halting problem /specifically/ asks about a machine simulating itself.None of above indicates any disagreement by any authority.Everyone (even Linz) has the wrong headed idea that a halt decider must report on the behavior of the computation that itself is contained
within. This has always been wrong.
A halt decider must always report on the behavior that its finite string specifies. This is different only when an input invokes its own decider.Um, no? Then it is making a mistake.
H is not an UTM, though.The one rare exception is shown above where Ĥ ⟨Ĥ⟩ halts and the input >>> to embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly reach its own final state of
⟨Ĥ.qn⟩ when embedded_H acts as if it was a UTM.
Which authority? Not that that would be a valid argument.That is not supported by any anuthority.The authority says *given an input of the function domain it*
*can return the corresponding output*
In other words all deciders compute the mapping from their input (finite string) to an accept or reject state.They do, if those happen to coincide.
This means that they do not compute the mapping of the executing process
of themselves.
I am the first person in the world that noticed these two could beWhat are the semantics that you disagree about?
different. Everyone that has disagreed with me is disagreeing with the semantics of the x86 language.
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