• Re: Correcting the definition of the terms of the halting problem[3]

    From Mikko@21:1/5 to olcott on Sat Jan 20 11:03:41 2024
    On 2024-01-19 16:55:15 +0000, olcott said:

    When Ĥ is applied to ⟨Ĥ⟩
    Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
    Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

    Although the second and third lines are good sentences with clear
    meanings, the pair looks more like random noise. Perhaps there should
    be a conjunction between them or something else to connect them to
    something that makes sense in the context.

    Mikko

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  • From Mikko@21:1/5 to Alan Mackenzie on Sat Jan 20 11:37:28 2024
    On 2024-01-19 22:50:34 +0000, Alan Mackenzie said:

    In comp.theory Richard Damon <[email protected]> wrote:

    [ .... ]

    Only a simulation that shows that you can not reach a final state even
    after an unbounded number of steps shows non-halting, but doing such a
    simulation make the machine fail to be a decider, as deciders must
    answer in bounded time, and until you invent a time machine, you can't
    do unbounded work in bounded time.

    Is that right? Deciders must answer in finite time, but is there
    actually a bound on how long this time can be? If so, what is it?

    Both the decider and every possible input to the decider can be
    described with a finite string. There are only a finite number of
    program-input pairs that can be described with a string of the same
    length or shorter. Each of those pairs represent a halting or
    non-halting computation. Each of the halting computation has a
    computation time, and because the number of those computations
    is finite, there is a maximum time that is the time of one or more
    of those computations. Considering all descriptions of program-input
    pairs of any length, for each length there is a maximum time that
    is only exceeded by non-halting programs. However, there is no
    way to compute that time from the length.

    Mikko

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