• Re: Olcott's H(P,P)==0 is correct !!!

    From olcott@21:1/5 to Richard Damon on Wed Jun 8 18:56:28 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have determined that
    his so called refutation of the HP proofs is based around the
    behaviour of his simulation-based decider, H:

    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being detected (i.e. Q >>>> invoking H) irregardless of whether FUBAR halts or not.

    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested) simulation >>>> however that wouldn't be the case for non-simulating decider for which >>>> there would be no such recursion.

    Can we finally put this to bed and change the fucking topic?

    /Flibble

    +-----------------------------------------------------------------+
    | The HP proof has nothing to do with how the 'H' is constructed. |
    +-----------------------------------------------------------------+
    Many such liar's-paradox-like examples are for easy comprehension
    (for educational purpose).
    The real 'H' inside P is an algorithm computationally equivalent to
    'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent god....
    whatever).


    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs halt, a
    "pathological"
          program P, called with some input, can pass its own source and
    its input to
          H and then specifically do the opposite of what H predicts P
    will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical operation,
    olcott cannot
    make it formally correct (he relies on his "language's logic").
    For example, I doubt he can handle several lines of inter-connected
    codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is
    correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt status of
    its input. I am claiming that non-halting is the correct halt status
    of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right answer,
    but H doesn't actually return that answer, you can be correct on that.


    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the "impossible"
    input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as they
    possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally obvious
    fact that H(P,P)==0 is correct. With an actual honest dialogue there is
    no way that this should have taken more than three days to get everyone
    to agree.

    As soon as we achieve a consensus on this point we can move on to the
    next point.

    --
    Copyright 2022 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 Mr Flibble@21:1/5 to olcott on Thu Jun 9 01:04:14 2022
    XPost: comp.theory, sci.logic, sci.math

    On Wed, 8 Jun 2022 18:56:28 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have determined
    that his so called refutation of the HP proofs is based around
    the behaviour of his simulation-based decider, H:

    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being detected
    (i.e. Q invoking H) irregardless of whether FUBAR halts or not.

    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested)
    simulation however that wouldn't be the case for non-simulating
    decider for which there would be no such recursion.

    Can we finally put this to bed and change the fucking topic?

    /Flibble

    +-----------------------------------------------------------------+
    | The HP proof has nothing to do with how the 'H' is constructed.
    |
    +-----------------------------------------------------------------+
    Many such liar's-paradox-like examples are for easy comprehension
    (for educational purpose). The real 'H' inside P is an algorithm
    computationally equivalent to 'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent
    god.... whatever).


    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs halt, a
    "pathological"
          program P, called with some input, can pass its own source
    and its input to
          H and then specifically do the opposite of what H predicts P
    will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical operation,
    olcott cannot
    make it formally correct (he relies on his "language's logic").
    For example, I doubt he can handle several lines of
    inter-connected codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is
    correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt status
    of its input. I am claiming that non-halting is the correct halt
    status of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right
    answer, but H doesn't actually return that answer, you can be
    correct on that.

    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the
    "impossible" input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as they possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally
    obvious fact that H(P,P)==0 is correct. With an actual honest
    dialogue there is no way that this should have taken more than three
    days to get everyone to agree.

    As soon as we achieve a consensus on this point we can move on to the
    next point.

    Your H is not a halt decider as it gets the answer wrong if P would
    have halted; instead what you have is an S, a simulation detector.

    So if you wish to continue you should be asserting S(P,P)==0 and not
    H(P,P)==0.

    /Flibble

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to Mr Flibble on Wed Jun 8 19:10:39 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/8/2022 7:04 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 18:56:28 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have determined
    that his so called refutation of the HP proofs is based around
    the behaviour of his simulation-based decider, H:

    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being detected
    (i.e. Q invoking H) irregardless of whether FUBAR halts or not.

    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested)
    simulation however that wouldn't be the case for non-simulating
    decider for which there would be no such recursion.

    Can we finally put this to bed and change the fucking topic?

    /Flibble

    +-----------------------------------------------------------------+
    | The HP proof has nothing to do with how the 'H' is constructed.
    |
    +-----------------------------------------------------------------+
    Many such liar's-paradox-like examples are for easy comprehension
    (for educational purpose). The real 'H' inside P is an algorithm
    computationally equivalent to 'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent
    god.... whatever).


    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs halt, a >>>> "pathological"
          program P, called with some input, can pass its own source >>>> and its input to
          H and then specifically do the opposite of what H predicts P >>>> will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical operation,
    olcott cannot
    make it formally correct (he relies on his "language's logic").
    For example, I doubt he can handle several lines of
    inter-connected codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is
    correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt status
    of its input. I am claiming that non-halting is the correct halt
    status of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right
    answer, but H doesn't actually return that answer, you can be
    correct on that.

    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the
    "impossible" input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as they
    possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally
    obvious fact that H(P,P)==0 is correct. With an actual honest
    dialogue there is no way that this should have taken more than three
    days to get everyone to agree.

    As soon as we achieve a consensus on this point we can move on to the
    next point.

    Your H is not a halt decider as it gets the answer wrong if P would
    have halted; instead what you have is an S, a simulation detector.

    So if you wish to continue you should be asserting S(P,P)==0 and not H(P,P)==0.

    /Flibble


    So the concept of a simulating halt decider is beyond your intellectual capacity?

    --
    Copyright 2022 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 Mr Flibble@21:1/5 to olcott on Thu Jun 9 01:13:01 2022
    XPost: comp.theory, sci.logic, sci.math

    On Wed, 8 Jun 2022 19:10:39 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 7:04 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 18:56:28 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have
    determined that his so called refutation of the HP proofs is
    based around the behaviour of his simulation-based decider, H:

    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being detected
    (i.e. Q invoking H) irregardless of whether FUBAR halts or not.

    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested)
    simulation however that wouldn't be the case for non-simulating
    decider for which there would be no such recursion.

    Can we finally put this to bed and change the fucking topic?

    /Flibble

    +-----------------------------------------------------------------+ >>>>> | The HP proof has nothing to do with how the 'H' is
    constructed. |
    +-----------------------------------------------------------------+ >>>>> Many such liar's-paradox-like examples are for easy
    comprehension (for educational purpose). The real 'H' inside P
    is an algorithm computationally equivalent to 'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent
    god.... whatever).


    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs halt, a
    "pathological"
          program P, called with some input, can pass its own source
    and its input to
          H and then specifically do the opposite of what H
    predicts P will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical operation,
    olcott cannot
    make it formally correct (he relies on his "language's logic").
    For example, I doubt he can handle several lines of
    inter-connected codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is
    correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt
    status of its input. I am claiming that non-halting is the
    correct halt status of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right
    answer, but H doesn't actually return that answer, you can be
    correct on that.

    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the
    "impossible" input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as
    they possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally
    obvious fact that H(P,P)==0 is correct. With an actual honest
    dialogue there is no way that this should have taken more than
    three days to get everyone to agree.

    As soon as we achieve a consensus on this point we can move on to
    the next point.

    Your H is not a halt decider as it gets the answer wrong if P would
    have halted; instead what you have is an S, a simulation detector.

    So if you wish to continue you should be asserting S(P,P)==0 and not H(P,P)==0.

    /Flibble


    So the concept of a simulating halt decider is beyond your
    intellectual capacity?

    It isn't a halt decider if it gets the answer to the question of
    whether or not P halts wrong. Your H gets the answer wrong if P halts
    so it isn't a halt decider.

    /Flibble

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From Richard Damon@21:1/5 to olcott on Wed Jun 8 20:16:34 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/8/22 7:56 PM, olcott wrote:
    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have determined that >>>>> his so called refutation of the HP proofs is based around the
    behaviour of his simulation-based decider, H:

    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being detected
    (i.e. Q
    invoking H) irregardless of whether FUBAR halts or not.

    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested) simulation >>>>> however that wouldn't be the case for non-simulating decider for which >>>>> there would be no such recursion.

    Can we finally put this to bed and change the fucking topic?

    /Flibble

    +-----------------------------------------------------------------+
    | The HP proof has nothing to do with how the 'H' is constructed. |
    +-----------------------------------------------------------------+
    Many such liar's-paradox-like examples are for easy comprehension
    (for educational purpose).
    The real 'H' inside P is an algorithm computationally equivalent to
    'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent
    god.... whatever).


    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs halt, a
    "pathological"
          program P, called with some input, can pass its own source and >>> its input to
          H and then specifically do the opposite of what H predicts P
    will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical operation,
    olcott cannot
    make it formally correct (he relies on his "language's logic").
    For example, I doubt he can handle several lines of inter-connected
    codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is
    correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt status of
    its input. I am claiming that non-halting is the correct halt status
    of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right answer,
    but H doesn't actually return that answer, you can be correct on that.


    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the "impossible"
    input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as they possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally obvious
    fact that H(P,P)==0 is correct. With an actual honest dialogue there is
    no way that this should have taken more than three days to get everyone
    to agree.

    As soon as we achieve a consensus on this point we can move on to the
    next point.


    The problem is you want to claim that not only is it the correct answer,
    but also that H(P,P) returns it.

    As I explained, the answer is correct only if H doesn't actually return it.

    If H(P,P) returns 0, then the correct answer will be 1, because P(P)
    will Halt.

    This is the key point in the Halting Theorem proof, by the construction
    of H^,

    if H applied to <H^> <H^> answers non-halting, then H^ applied to <H^>
    will Halt, and thus the answer that H gave will be wrong.

    If H applied to <H^> <H^> answers halting, then H^ applied to <H^> will
    go into an infinite loop, and thus the answer that H gave will be wrong.

    If H applied to <H^> <H^> does any other action (never halting or
    halting in some other state), H is just wrong for not meeting the basic requirements as a decider.

    Thus, there is NO behavior that H applied to <H^> <H^> can do that is
    correct, thus there can not be an H that is correct for all inputs,
    since <H^> <H^> is a legal input, and demonstrating even ONE counter
    example disproves an "For ALL" requirement.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to Mr Flibble on Wed Jun 8 19:17:11 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/8/2022 7:13 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 19:10:39 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 7:04 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 18:56:28 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have
    determined that his so called refutation of the HP proofs is
    based around the behaviour of his simulation-based decider, H: >>>>>>>>
    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being detected >>>>>>>> (i.e. Q invoking H) irregardless of whether FUBAR halts or not. >>>>>>>>
    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested)
    simulation however that wouldn't be the case for non-simulating >>>>>>>> decider for which there would be no such recursion.

    Can we finally put this to bed and change the fucking topic?

    /Flibble

    +-----------------------------------------------------------------+ >>>>>>> | The HP proof has nothing to do with how the 'H' is
    constructed. |
    +-----------------------------------------------------------------+ >>>>>>> Many such liar's-paradox-like examples are for easy
    comprehension (for educational purpose). The real 'H' inside P
    is an algorithm computationally equivalent to 'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent
    god.... whatever).


    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs halt, a >>>>>> "pathological"
          program P, called with some input, can pass its own source >>>>>> and its input to
          H and then specifically do the opposite of what H
    predicts P will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical operation, >>>>>>> olcott cannot
    make it formally correct (he relies on his "language's logic").
    For example, I doubt he can handle several lines of
    inter-connected codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is
    correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt
    status of its input. I am claiming that non-halting is the
    correct halt status of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right
    answer, but H doesn't actually return that answer, you can be
    correct on that.

    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the
    "impossible" input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as
    they possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally
    obvious fact that H(P,P)==0 is correct. With an actual honest
    dialogue there is no way that this should have taken more than
    three days to get everyone to agree.

    As soon as we achieve a consensus on this point we can move on to
    the next point.

    Your H is not a halt decider as it gets the answer wrong if P would
    have halted; instead what you have is an S, a simulation detector.

    So if you wish to continue you should be asserting S(P,P)==0 and not
    H(P,P)==0.

    /Flibble


    So the concept of a simulating halt decider is beyond your
    intellectual capacity?

    It isn't a halt decider if it gets the answer to the question of
    whether or not P halts wrong. Your H gets the answer wrong if P halts
    so it isn't a halt decider.

    /Flibble



    (1) Deciders(computer science) compute the mapping from their inputs to
    an accept or reject state.

    (2) The actual behavior of the actual input to H(P,P) is proven to never
    halt.

    (3) P(P) is not an input to H, thus out-of-scope for H.

    The actual behavior of the correctly simulated input to H(P,P) is not
    the same as the actual behavior of P(P) because the specified sequence
    of instructions is not the same.

    Because halt deciders must report on the actual behavior of their actual
    inputs H(P,P)==0 is correct even though P(P) halts.

    int sum(int x, int y)
    {
    return x + y;
    }

    Expecting H(P,P) to report on the behavior of an entirely different
    sequence of instructions than its input actually specifies is exactly
    the same as expecting sum(3,4) to return the sum of 5 + 7, quite nuts.

    --
    Copyright 2022 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 Mr Flibble@21:1/5 to olcott on Thu Jun 9 01:19:29 2022
    XPost: comp.theory, sci.logic, sci.math

    On Wed, 8 Jun 2022 19:17:11 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 7:13 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 19:10:39 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 7:04 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 18:56:28 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble
    wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have
    determined that his so called refutation of the HP proofs is >>>>>>>> based around the behaviour of his simulation-based decider,
    H:

    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being
    detected (i.e. Q invoking H) irregardless of whether FUBAR
    halts or not.

    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested)
    simulation however that wouldn't be the case for
    non-simulating decider for which there would be no such
    recursion.

    Can we finally put this to bed and change the fucking topic? >>>>>>>>
    /Flibble

    +-----------------------------------------------------------------+ >>>>>>> | The HP proof has nothing to do with how the 'H' is
    constructed. |
    +-----------------------------------------------------------------+ >>>>>>> Many such liar's-paradox-like examples are for easy
    comprehension (for educational purpose). The real 'H' inside P >>>>>>> is an algorithm computationally equivalent to 'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent
    god.... whatever).


    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs
    halt, a "pathological"
          program P, called with some input, can pass its own
    source and its input to
          H and then specifically do the opposite of what H
    predicts P will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical
    operation, olcott cannot
    make it formally correct (he relies on his "language's
    logic"). For example, I doubt he can handle several lines of
    inter-connected codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is >>>>>>> correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt
    status of its input. I am claiming that non-halting is the
    correct halt status of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right
    answer, but H doesn't actually return that answer, you can be
    correct on that.

    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the
    "impossible" input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as
    they possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally
    obvious fact that H(P,P)==0 is correct. With an actual honest
    dialogue there is no way that this should have taken more than
    three days to get everyone to agree.

    As soon as we achieve a consensus on this point we can move on to
    the next point.

    Your H is not a halt decider as it gets the answer wrong if P
    would have halted; instead what you have is an S, a simulation
    detector.

    So if you wish to continue you should be asserting S(P,P)==0 and
    not H(P,P)==0.

    /Flibble


    So the concept of a simulating halt decider is beyond your
    intellectual capacity?

    It isn't a halt decider if it gets the answer to the question of
    whether or not P halts wrong. Your H gets the answer wrong if P
    halts so it isn't a halt decider.

    /Flibble



    (1) Deciders(computer science) compute the mapping from their inputs
    to an accept or reject state.

    (2) The actual behavior of the actual input to H(P,P) is proven to
    never halt.

    (3) P(P) is not an input to H, thus out-of-scope for H.

    The actual behavior of the correctly simulated input to H(P,P) is not
    the same as the actual behavior of P(P) because the specified
    sequence of instructions is not the same.

    Because halt deciders must report on the actual behavior of their
    actual inputs H(P,P)==0 is correct even though P(P) halts.

    int sum(int x, int y)
    {
    return x + y;
    }

    Expecting H(P,P) to report on the behavior of an entirely different
    sequence of instructions than its input actually specifies is exactly
    the same as expecting sum(3,4) to return the sum of 5 + 7, quite nuts.

    You just keep repeating yourself; I can also repeat myself.

    Your H is not a halt decider as it gets the answer wrong if P would
    have halted; instead what you have is an S, a simulation detector.

    /Flibble

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From olcott@21:1/5 to Mr Flibble on Wed Jun 8 19:32:45 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/8/2022 7:19 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 19:17:11 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 7:13 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 19:10:39 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 7:04 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 18:56:28 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble
    wrote:
    Hi,

     From discussion with Olcott in comp.lang.c++ I have
    determined that his so called refutation of the HP proofs is >>>>>>>>>> based around the behaviour of his simulation-based decider, >>>>>>>>>> H:

    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being
    detected (i.e. Q invoking H) irregardless of whether FUBAR >>>>>>>>>> halts or not.

    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested) >>>>>>>>>> simulation however that wouldn't be the case for
    non-simulating decider for which there would be no such
    recursion.

    Can we finally put this to bed and change the fucking topic? >>>>>>>>>>
    /Flibble

    +-----------------------------------------------------------------+ >>>>>>>>> | The HP proof has nothing to do with how the 'H' is
    constructed. |
    +-----------------------------------------------------------------+ >>>>>>>>> Many such liar's-paradox-like examples are for easy
    comprehension (for educational purpose). The real 'H' inside P >>>>>>>>> is an algorithm computationally equivalent to 'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent >>>>>>>>> god.... whatever).


    It is the pathological self reference(Olcott 2004) relationship >>>>>>>> between H and P that has previously been considered to make P
    undecidable for H.

          For any program H that might determine if programs >>>>>>>> halt, a "pathological"
          program P, called with some input, can pass its own >>>>>>>> source and its input to
          H and then specifically do the opposite of what H >>>>>>>> predicts P will do. No H
          can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem

    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical
    operation, olcott cannot
    make it formally correct (he relies on his "language's
    logic"). For example, I doubt he can handle several lines of >>>>>>>>> inter-connected codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is >>>>>>>>> correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt
    status of its input. I am claiming that non-halting is the
    correct halt status of its input.


    Ok, so if you are JUST claiming that Non-Halting is the right
    answer, but H doesn't actually return that answer, you can be
    correct on that.

    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the
    "impossible" input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as
    they possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally
    obvious fact that H(P,P)==0 is correct. With an actual honest
    dialogue there is no way that this should have taken more than
    three days to get everyone to agree.

    As soon as we achieve a consensus on this point we can move on to
    the next point.

    Your H is not a halt decider as it gets the answer wrong if P
    would have halted; instead what you have is an S, a simulation
    detector.

    So if you wish to continue you should be asserting S(P,P)==0 and
    not H(P,P)==0.

    /Flibble


    So the concept of a simulating halt decider is beyond your
    intellectual capacity?

    It isn't a halt decider if it gets the answer to the question of
    whether or not P halts wrong. Your H gets the answer wrong if P
    halts so it isn't a halt decider.

    /Flibble



    (1) Deciders(computer science) compute the mapping from their inputs
    to an accept or reject state.

    (2) The actual behavior of the actual input to H(P,P) is proven to
    never halt.

    (3) P(P) is not an input to H, thus out-of-scope for H.

    The actual behavior of the correctly simulated input to H(P,P) is not
    the same as the actual behavior of P(P) because the specified
    sequence of instructions is not the same.

    Because halt deciders must report on the actual behavior of their
    actual inputs H(P,P)==0 is correct even though P(P) halts.

    int sum(int x, int y)
    {
    return x + y;
    }

    Expecting H(P,P) to report on the behavior of an entirely different
    sequence of instructions than its input actually specifies is exactly
    the same as expecting sum(3,4) to return the sum of 5 + 7, quite nuts.

    You just keep repeating yourself; I can also repeat myself.

    Your H is not a halt decider as it gets the answer wrong if P would
    have halted;

    I have already proved that the correctly emulated input to H(P,P) would
    never halt and I did this at least 50 times.


    instead what you have is an S, a simulation detector.

    /Flibble



    --
    Copyright 2022 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 Richard Damon@21:1/5 to olcott on Wed Jun 8 20:46:10 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/8/22 8:17 PM, olcott wrote:
    On 6/8/2022 7:13 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 19:10:39 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/8/2022 7:04 PM, Mr Flibble wrote:
    On Wed, 8 Jun 2022 18:56:28 -0500
    olcott <NoOne@NoWhere.com> wrote:
    On 6/8/2022 6:37 PM, Richard Damon wrote:
    On 6/8/22 11:53 AM, olcott wrote:
    On 6/8/2022 10:37 AM, wij wrote:
    On Wednesday, 8 June 2022 at 08:43:00 UTC+8, Mr Flibble wrote: >>>>>>>>> Hi,

       From discussion with Olcott in comp.lang.c++ I have
    determined that his so called refutation of the HP proofs is >>>>>>>>> based around the behaviour of his simulation-based decider, H: >>>>>>>>>
    void Q(u32 x)
    {
    if (H(x, x))
    FUBAR();
    return;
    }

    int main()
    {
    Output("Input_Halts = ", H((u32)Q, (u32)Q));
    }

    He asserts H(Q,Q)=0 based on a nested simulation being detected >>>>>>>>> (i.e. Q invoking H) irregardless of whether FUBAR halts or not. >>>>>>>>>
    If FUBAR halts H gives the wrong answer.

    He claims H(Q,Q)=0 as it gets stuck in a recursive (nested)
    simulation however that wouldn't be the case for non-simulating >>>>>>>>> decider for which there would be no such recursion.

    Can we finally put this to bed and change the fucking topic? >>>>>>>>>
    /Flibble

    +-----------------------------------------------------------------+ >>>>>>>> | The HP proof has nothing to do with how the 'H' is
    constructed. |
    +-----------------------------------------------------------------+ >>>>>>>> Many such liar's-paradox-like examples are for easy
    comprehension (for educational purpose). The real 'H' inside P >>>>>>>> is an algorithm computationally equivalent to 'H' (so, no
    any 'call' is there, and the pattern matching tech. is very
    difficult, by now to say.
    And, this 'H' is also allowed given by some super intelligent
    god.... whatever).

    It is the pathological self reference(Olcott 2004) relationship
    between H and P that has previously been considered to make P
    undecidable for H.

            For any program H that might determine if programs halt, a
    "pathological"
            program P, called with some input, can pass its own source
    and its input to
            H and then specifically do the opposite of what H
    predicts P will do. No H
            can exist that handles this case.
    https://en.wikipedia.org/wiki/Halting_problem
    +------------------------------------------------+
    | olcott's brain is incapable of logic function. |
    | (other kind of functions seem quite excellent) |
    +------------------------------------------------+
    It should be safe to say any concept involves logical operation, >>>>>>>> olcott cannot
    make it formally correct (he relies on his "language's logic"). >>>>>>>> For example, I doubt he can handle several lines of
    inter-connected codes.
    ...
    All should be fine... except olcott now insists "H(P,P)==0" is >>>>>>>> correct while
    there is no definition of H shown.

    I am not claiming that H(P,P) correctly determines the halt
    status of its input. I am claiming that non-halting is the
    correct halt status of its input.

    Ok, so if you are JUST claiming that Non-Halting is the right
    answer, but H doesn't actually return that answer, you can be
    correct on that.

    THIS BY ITSELF IS AN ENORMOUS BREAKTHROUGH
    I am claiming that H(P,P)==0 is the correct answer to the
    "impossible" input that previously had no correct answer at all.

    Because people on this forum are trying to be as disagreeable as
    they possibly can be I must move one point at a time.

    It has taken at least six months to get agreement on the totally
    obvious fact that H(P,P)==0 is correct. With an actual honest
    dialogue there is no way that this should have taken more than
    three days to get everyone to agree.

    As soon as we achieve a consensus on this point we can move on to
    the next point.
    Your H is not a halt decider as it gets the answer wrong if P would
    have halted; instead what you have is an S, a simulation detector.

    So if you wish to continue you should be asserting S(P,P)==0 and not
    H(P,P)==0.

    /Flibble

    So the concept of a simulating halt decider is beyond your
    intellectual capacity?
    It isn't a halt decider if it gets the answer to the question of
    whether or not P halts wrong.  Your H gets the answer wrong if P halts
    so it isn't a halt decider.

    /Flibble



    (1) Deciders(computer science) compute the mapping from their inputs to
    an accept or reject state.

    (2) The actual behavior of the actual input to H(P,P) is proven to never halt.

    ONLY if H(P,P) never returns 0.


    (3) P(P) is not an input to H, thus out-of-scope for H.

    WRONG. Shows you don't understand deciders.

    The input to H is the desciption of P(P), (or H just isn't a Halt
    Decider), and the JOB of H is thus to determine if the computation that
    input represents will halt or not.



    The actual behavior of the correctly simulated input to H(P,P) is not
    the same as the actual behavior of P(P) because the specified sequence
    of instructions is not the same.

    WRONG, by the definition of "Correct Simulation" and the definition of a
    Halt Decider, the "Correct Simulation" of the input to a Halt Decider is
    in fact the behavior of the computation the input represents.

    WHAT is the first instruction that the sequence of instructions diverge?


    Because halt deciders must report on the actual behavior of their actual inputs H(P,P)==0 is correct even though P(P) halts.


    WRONG, BY DEFINITION.

    int sum(int x, int y)
    {
      return x + y;
    }

    Expecting H(P,P) to report on the behavior of an entirely different
    sequence of instructions than its input actually specifies is exactly
    the same as expecting sum(3,4) to return the sum of 5 + 7, quite nuts.


    Nope, shows your stupidity, since the DEFINITION of the behavior of the
    input for a Halt Decider is the behavior of the computation the input represents.

    Your claim is like saying that cats bark because you dog, you call cat,
    barks.

    The fact you use wrong definitions doesn't mean you get to change that
    actual meaning of the words in the Theorems.

    Just shows you are a patholgical liar or a total idiot.

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