• Re: Simulating halt deciders correct decider halting [ Key error ][ bra

    From olcott@21:1/5 to Richard Damon on Thu Mar 17 19:20:43 2022
    XPost: comp.theory, sci.logic, sci.math

    On 3/17/2022 5:30 PM, Richard Damon wrote:

    On 3/17/22 5:30 PM, olcott wrote:
    On 3/17/2022 3:36 PM, Richard Damon wrote:

    On 3/17/22 4:00 PM, olcott wrote:
    On 3/17/2022 2:30 PM, Richard Damon wrote:

    On 3/17/22 10:06 AM, olcott wrote:
    On 3/16/2022 10:02 PM, Richard Damon wrote:
    Ĥ applied to ⟨Ĥ⟩ reaches its final state only when embedded_H >>>>>> applied to ⟨Ĥ⟩ ⟨Ĥ⟩ aborts its simulation.

    Right, but it DOES reach its final state, so embedded_H, to be
    correct, must take that into account. it doesn't, so it gives the
    WRONG answer.


    No BRAIN DEAD MORON the simulated ⟨Ĥ⟩ never reaches its own final >>>> state.


    The CORRECT Simulation shows that it does.

    H's simulation only doesn't reach that final state because H aborted
    it simulation.


    We are not talking about the simulation reaching the final state of
    the simulation BRAIN DEAD MORON we are talking about the simulated
    input reaching its own final state.

    When Ĥ is applied to ⟨Ĥ⟩
       Ĥ copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩

    Then these steps would keep repeating:
       Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
       Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩
       Ĥ3 copies its input ⟨Ĥ4⟩ to ⟨Ĥ5⟩ then embedded_H simulates ⟨Ĥ4⟩
    ⟨Ĥ5⟩...

    The simulated input never reaches its own final state whether or not
    embedded_H aborts this simulation.


    No, because if embedded_H aborts its simulation, then the trace must be:

    When H^ is applied <H^>
    H^0 copies its input <H^1> to <H^2> then H0 simulates <H^1> <H^2>
    H^1 copies its input <H^2> to <H^3> then H1 simulates <H^2> <H^3>
    H^2 copies its input <H^3> to <H^4> then H2 simulates <H^3> <H^4>
    ... n levels (to when H0 aborts
    H0 goes to H0.Qn and H^0 HALTS.


    Incorrect notation.

    When Ĥ is applied to ⟨Ĥ⟩
    Ĥ copies its input ⟨Ĥ0⟩ to ⟨Ĥ1⟩ then embedded_H simulates ⟨Ĥ0⟩ ⟨Ĥ1⟩

    Then these steps would keep repeating:
    Ĥ0 copies its input ⟨Ĥ1⟩ to ⟨Ĥ2⟩ then embedded_H0 simulates ⟨Ĥ1⟩ ⟨Ĥ2⟩...
    Ĥ1 copies its input ⟨Ĥ2⟩ to ⟨Ĥ3⟩ then embedded_H1 simulates ⟨Ĥ2⟩ ⟨Ĥ3⟩
    Ĥ2 copies its input ⟨Ĥ3⟩ to ⟨Ĥ4⟩ then embedded_H2 simulates ⟨Ĥ3⟩ ⟨Ĥ4⟩

    When the original executing (not simulated) embedded_H sees the
    infinitely repeating pattern then it aborts its simulation killing every simulation in the whole process tree.


    I.E. if embedded_H will only simulate a finite number of iterations, and
    them abort its simulation, there is NO infinite simulation that happens.

    No BRAIN DEAD MORON the question is:
    Can the simulated ⟨Ĥ⟩ possibly ever reach its own final state?
    (a) embedded_H does not abort its simulation: NO
    (b) embedded_H aborts its simulation: NO

    It is all covered by the self-evidently correct general principle:

    When-so-ever the simulation of any input to a simulating halt decider
    must be aborted to prevent its infinite simulation this input is
    correctly rejected as non-halting.



    --
    Copyright 2021 Pete Olcott

    Talent hits a target no one else can hit;
    Genius hits a target no one else can see.
    Arthur Schopenhauer

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to Ben Bacarisse on Fri Mar 18 12:41:41 2022
    XPost: comp.theory, sci.logic, sci.math

    On 3/18/2022 11:54 AM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    No BRAIN DEAD MORON this is not true, I keep calling you a BRAIN DEAD
    MORON because after I have explain all the details you cannot remember
    what I just said.

    I see the petulant six-year-old is in residence today.

    A decider maps its inputs to its own accept reject state.
    A halt decider does not compute the halt status of itself.

    A halt decider, let's call it H, maps the input <Ĥ> <Ĥ> to its accept
    state if (and only if) Ĥ enter a final state on input <Ĥ>,

    No that is entirely incorrect.

    embedded_H maps its input <Ĥ> <Ĥ> to its Ĥ.qn final reject state if the correctly simulated input <Ĥ> <Ĥ> would never reach its own final state
    of <Ĥ>.qn in any finite number of steps of simulation.


    and it maps
    <Ĥ> <Ĥ> to its reject state if (and only if) Ĥ does not enter a final
    state on input <Ĥ>.

    No such H exists.

    You implicitly accept this fact because you propose an H which does not
    meet this specification but I can't see why you think anyone would care
    about it.



    --
    Copyright 2021 Pete Olcott

    Talent hits a target no one else can hit;
    Genius hits a target no one else can see.
    Arthur Schopenhauer

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to Richard Damon on Fri Mar 18 13:28:39 2022
    XPost: comp.theory, sci.logic, sci.math

    On 3/18/2022 12:55 PM, Richard Damon wrote:
    On 3/18/22 1:41 PM, olcott wrote:
    On 3/18/2022 11:54 AM, Ben Bacarisse wrote:
    olcott <NoOne@NoWhere.com> writes:

    No BRAIN DEAD MORON this is not true, I keep calling you a BRAIN DEAD
    MORON because after I have explain all the details you cannot remember >>>> what I just said.

    I see the petulant six-year-old is in residence today.

    A decider maps its inputs to its own accept reject state.
    A halt decider does not compute the halt status of itself.

    A halt decider, let's call it H, maps the input <Ĥ> <Ĥ> to its accept
    state if (and only if) Ĥ enter a final state on input <Ĥ>,

    No that is entirely incorrect.

    It the DEFINITION!!!!


    embedded_H maps its input <Ĥ> <Ĥ> to its Ĥ.qn final reject state if
    the correctly simulated input <Ĥ> <Ĥ> would never reach its own final
    state of <Ĥ>.qn in any finite number of steps of simulation.

    WRONG CRITERIA.

    A decider computes the mapping from its input to its own accept or
    reject state.

    A halt decider computes the mapping from its input to its own accept or
    reject state on the basis of the behavior specified by its input.

    Since a simulating halt decider has a complete UTM as a part of its own functionality it can examine the behavior of the correct pure simulation
    of N steps of simulation.

    If the input specifies a halting computation then the simulation will
    complete in N steps.

    If the input specifies a non halting sequence of configurations then the simulation would never complete in any finite number of steps.

    If simulation would never complete in any finite number of steps, then
    it would be necessarily correct for the simulating halt decider to
    reject this input.

    That is just your POOP which no one cares about. We care
    about the simulation by a REAL UTM, which matches the actual behavior of
    the Turing Machine / Input combination represented by the input, not the behavior that H sees in its partial simulation.



    and it maps
    <Ĥ> <Ĥ> to its reject state if (and only if) Ĥ does not enter a final >>> state on input <Ĥ>.

    No such H exists.

    You implicitly accept this fact because you propose an H which does not
    meet this specification but I can't see why you think anyone would care
    about it.






    --
    Copyright 2021 Pete Olcott

    Talent hits a target no one else can hit;
    Genius hits a target no one else can see.
    Arthur Schopenhauer

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)