• Re: Halting Problem solved! [no mistake]

    From olcott@21:1/5 to Mr Flibble on Sun Apr 23 18:30:22 2023
    XPost: comp.theory, sci.logic, sci.math
    XPost: alt.philosophy

    On 4/23/2023 5:26 PM, Mr Flibble wrote:
    On 23/04/2023 10:23 pm, Richard Damon wrote:
    On 4/23/23 4:28 PM, Mr Flibble wrote:
    On 23/04/2023 6:16 pm, Richard Damon wrote:
    On 4/23/23 1:06 PM, Mr Flibble wrote:
    On 23/04/2023 5:44 pm, Richard Damon wrote:
    On 4/23/23 8:47 AM, Mr Flibble wrote:
    On 23/04/2023 12:27 pm, Richard Damon wrote:
    On 4/22/23 11:47 PM, Mr Flibble wrote:
    On 23/04/2023 3:42 am, Richard Damon wrote:
    On 4/22/23 9:55 PM, Mr Flibble wrote:
    On 23/04/2023 1:39 am, Richard Damon wrote:
    On 4/22/23 8:31 PM, Mr Flibble wrote:
    On 23/04/2023 12:53 am, Richard Damon wrote:
    On 4/22/23 7:43 PM, Mr Flibble wrote:
    Hi!

    I have an idea for a signaling simulating halt decider >>>>>>>>>>>>>>> that forks the
    simulation into two branches if the input calls the halt >>>>>>>>>>>>>>> decider as
    per [Strachey 1965]'s "Impossible Program":

    void P(void (*x)())
    {
         if (H(x, x))
             infinite_loop: goto infinite_loop; >>>>>>>>>>>>>>>      return;
    }

    int main()
    {
         std::cout << "Input halts: " << H(P, P) << std::endl; >>>>>>>>>>>>>>> }

    When the simulator detects the call to H in P it forks >>>>>>>>>>>>>>> the simulation
    into a non-halting branch (returning 0 to P) and a >>>>>>>>>>>>>>> halting branch
    (returning 1 to P) and continues the simulation of these >>>>>>>>>>>>>>> two branches
    in parallel.

    If the non-halting branch is determined to halt AND the >>>>>>>>>>>>>>> halting branch
    is determined to not halt then pathology is detected and >>>>>>>>>>>>>>> reported via
    a sNaP (signaling Not a Program) signal (analogous to >>>>>>>>>>>>>>> IEEE 754's
    sNaN (signaling Not a Number) signal)

    If EITHER branch is determined to be correctly decided >>>>>>>>>>>>>>> then that will
    be the decision of the halting decider.

    Crucially this scheme will handle (and correctly decide) the >>>>>>>>>>>>>>> following case whereby the result of H is discarded by >>>>>>>>>>>>>>> the input:

    void Px(void (*x)())
    {
         (void) H(x, x);
         return;
    }

    Obviously my idea necessitates extending the definition >>>>>>>>>>>>>>> of a halt
    decider:

    1) Decider decision is HALTS if input halts.
    2) Decider decision is NON-HALTING if input does not halt. >>>>>>>>>>>>>>> 3) Decider rejects pathological input as invalid by >>>>>>>>>>>>>>> signaling sNaP.

    Thoughts?  I am probably missing something obvious as my >>>>>>>>>>>>>>> idea
    appears to refute [Strachey 1965] and associated HP >>>>>>>>>>>>>>> proofs which
    great minds have mulled over for decades.

    /Flibble

    So, see if you can show an actual use for your altered >>>>>>>>>>>>>> definition of Halt Decoding.

    It will decide that P() is pathological input and it will >>>>>>>>>>>>> decide that Px() is halting.


    But those are just toy programs (P was just a simple program >>>>>>>>>>>> to show classical halting to not be useful)

    What USEFUL resutls can be gotten with your decider. Based >>>>>>>>>>>> on the following answers, its hard to see one.


    You also need to clarify the rules of you computation >>>>>>>>>>>>>> system, as you have previously commented that it can't >>>>>>>>>>>>>> obey the "normal" rules used in computability theory. >>>>>>>>>>>>>
    I believe you are referring to the fact that the halt >>>>>>>>>>>>> decider function and the intrinsic H(...) are a property of >>>>>>>>>>>>> the machine itself, H is much like the "move tape left" >>>>>>>>>>>>> function of a Turing Machine.  The only thing "abnormal" >>>>>>>>>>>>> about it is that such a function is not included in the >>>>>>>>>>>>> traditional definition of a Turing Machine.

    Your whole model of computation is at significant variance >>>>>>>>>>>> from the classical theoretical model.



    Also, how does your decider determine if the branch it is >>>>>>>>>>>>>> following is non-halting.

    The way any simulating halt decider would: through the >>>>>>>>>>>>> detection of repeated state given the machine and its >>>>>>>>>>>>> resources are finite in size.

    So only able to detect non-halting in machines goig into >>>>>>>>>>>> repeating loops, and not just that the computation keeps >>>>>>>>>>>> growing unbounded.

    Repeated state means a duplicate hash of the machine's finite >>>>>>>>>>> state.

    But not suitable for things like the Twin Primes problem or >>>>>>>>>> Collatz Conjecture.

    Most of the interesting problems don't end up in a simple
    infinite loop, but a loop counting through numbers that will >>>>>>>>>> never reach there terminal condition.
       >

    A very small set of the problems of normal interest in the >>>>>>>>>>>> theory.

    The size of the set is relative.  You are missing the point: >>>>>>>>>>> to be computable the machine's resources can not be
    unbounded. Only problems that are computable using the
    technology of the present era are of interest: one has to be >>>>>>>>>>> a pragmatist.

    /Flibble

    For many of the problems, the "limited" memory of the modern >>>>>>>>>> computer is unlikely to be the major limit. The "Very small >>>>>>>>>> set" was the number of problems that can be handled, not the >>>>>>>>>> physical size of the problems.

    Remember, the problems that Halting was designed for were
    things that a person with paper and pencil were trying to
    solve. Detecting "simple" loops wasn't the problem.

    I am not sure why you are equating repeated finite state with >>>>>>>>> "simple" loops.

    /Flibble

    Because your "repeated state" method won't catch even fairly
    simple issues like:

    for(i = 100; i != 1; i -= 2;) {
    // do something but don't change i
    }

    because the "state" never repeats

    Of course that state repeats (and will be caught): the integer
    "i" is of finite size so it will eventually wrap around to have
    the same bit pattern a second time.

    /Flibble

    Depends on its type. It could be a big int (infinite precision
    integer), so it runs until the machine exhausts its memory.

    If you are admitting that you can only handle "finite" machines,
    then that has been a solved problem for a long time. Even the
    pathological program can be solved under finite memory limits (it
    will reach memory exhaustion), which of course needs to be a
    fourth response possible out of your decider.

    Agree. As I posted earlier one has to be pragmatic given the finite
    constraints: a halt decision may not be possible on Machine A but
    may be possible on Machine B which has twice the resources of
    Machine A, for example.

    /Flibble

    Yep, well known answer to the Halting Problem for fixed sized
    machines, is a machine with (slightly more than) twice the memory
    needed for the program to decide, and run two copies of it, one at
    half speed.

    You compare the memories of the machines, and if the algorithm gets
    in a loop of length N, somewhere in 2N cycles of the faster machine,
    the two machines will line up and you will detect the repeated state.

    Yes, that sounds like a good solution and is what I would do if I was
    to actually implement the Flibble SSHD.

    /Flibble



    But that never generates your NaP exception, and never needs to, so
    the Flibble SSHD is shown to not be needed at all.

    The problem space being solved by it was already solved.

    Once you limit yourself to memory limited machines, the Halt Decider
    just needs to make sure that the "pathological" programs die of
    out-of-memory errors.

    No, the pathological program of [Strachey 1965] still needs to be
    explicitly detected as it won't die of an out-of-memory error: the so
    called "infinite recursion" of Olcott's decider is a mistake.

    /Flibble

    My C/x86 program is hypothesized to be a proxy for the behavior of an equivalent Turing machine.

    The idea was to understand the algorithm from the concrete X/x86 example
    and then apply the same reasoning to the Peter Linz Turing Machine based
    proof.

    My new paper reverses the order of this because so many people rejected
    the C/x86 out-of-hand as inapplicable to Turing Machines.

    *Simulating (partial) Halt Deciders Defeat the Halting Problem Proofs* https://www.researchgate.net/publication/369971402_Simulating_partial_Halt_Deciders_Defeat_the_Halting_Problem_Proofs

    --
    Copyright 2023 Olcott "Talent hits a target no one else can hit; Genius
    hits a target no one else can see." Arthur Schopenhauer

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