• Re: Experts would agree that my reviewers are incorrect [ deliberately

    From olcott@21:1/5 to Malcolm McLean on Tue May 31 17:23:11 2022
    XPost: comp.theory, sci.logic, sci.math

    On 5/31/2022 4:05 PM, Malcolm McLean wrote:
    On Tuesday, 31 May 2022 at 13:35:24 UTC+1, Malcolm McLean wrote:
    On Monday, 30 May 2022 at 15:35:38 UTC+1, olcott wrote:

    When we assume a simulating halt decider then none of the simulated
    inputs to the conventional proofs ever reach their final state when
    correctly simulated by this simulating halt decider.

    That you don't understand this does not make it false.

    We can't "assume a simulating halt decider" because the proof shows,
    or at least claims to show, that a simulating halt decider can't exist.

    None of the proofs bother to specifically examine simulating halt
    deciders this is their big mistake. All of the proofs simply assume that
    the execution trace of the input to H(P,P) reaches the
    self-contradictory part. This is a provably false assumption.

    --
    Copyright 2022 Pete 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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  • From Richard Damon@21:1/5 to olcott on Tue May 31 18:50:15 2022
    XPost: comp.theory, sci.logic, sci.math

    On 5/31/22 6:23 PM, olcott wrote:
    On 5/31/2022 4:05 PM, Malcolm McLean wrote:
    On Tuesday, 31 May 2022 at 13:35:24 UTC+1, Malcolm McLean wrote:
    On Monday, 30 May 2022 at 15:35:38 UTC+1, olcott wrote:

    When we assume a simulating halt decider then none of the simulated
    inputs to the conventional proofs ever reach their final state when
    correctly simulated by this simulating halt decider.

    That you don't understand this does not make it false.

    We can't "assume a simulating halt decider" because the proof shows,
    or at least claims to show, that a simulating halt decider can't exist.

    None of the proofs bother to specifically examine simulating halt
    deciders this is their big mistake. All of the proofs simply assume that
    the execution trace of the input to H(P,P) reaches the
    self-contradictory part. This is a provably false assumption.


    Because a simulating halt decider can't do something that NO halt
    decider can do.

    Your "Simulating Halt Decider" only gives what you call a correct answer because it isn't actually a Halt Decider, bcuase it isn't computing the
    Mapping of the Halting Problem.

    Your H isn't a Halt Decider, but a POOP decider.

    BY DEFINITION, A Halt Decider Accepts representations for ALL
    Machine/Input Combinations that Halt, and Rejects representations for
    ALL Machine/Inputs Combinations that never halt.

    Since H^ applied to H^ (ala P(P)) Halts when H is given there
    representation and answers that its input represents a Non-Halting
    Computation, H is just WRONG as a Halt Decider.

    If you can't represent that input to H, then it still fails, as you need
    to be able to represent ALL Machine/Input combinations.

    This is DEFINITIONALLY true.

    Note, the definition NEVER mentions an "Executiom Trace", that is only a transform allowed when the definition of simulation matches that of a
    UTM where the simulation DOES match the direct execution.

    Since you claim this isn't true of your simulation, you can't use that transform, and so you are shown to NOT be working on the Halting Problem.

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  • From olcott@21:1/5 to Malcolm McLean on Thu Jun 2 07:01:53 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 6:19 AM, Malcolm McLean wrote:
    On Wednesday, 1 June 2022 at 20:03:43 UTC+1, Ben wrote:
    Malcolm McLean <malcolm.ar...@gmail.com> writes:

    And actually, what you have done is construct a simulating attempted halt >>> decider, which fails on H_Hat, for reasons unrelated to the invert
    loop.
    What? H_Hat(H_Hat) halts because H(H_Hat, H_Hat) == 0 so, yes, no loop
    is involved, but that's not interesting.
    So you have actually achieved something of interest, if you could only
    recognise that.
    What do you think is new or interesting here? I can't see anything new
    or interesting here at all.

    The "simulating halt decider" will fail when fed H_Hat, not because of the invert
    step, but because it can't solve the simulation of H. That's inherent in the way
    it is set up - it can only terminate if it detects that the series of nested simulations is non-terminating, in which case it becomes terminating.


    That is not the point. The simulated input to H(P,P) only halts if it
    reaches its "ret" instruction otherwise it is non-halting.

    P(P) does halt yet it is an entirely different sequence of x86
    instructions than the correctly simulated input to H(P,P).

    People here say that they really really don't believe this thus directly disagree with the x86 language. This is a move that only one person here
    is dumb enough to make, yet more than one person does.

    I don't think anyone has actually pointed that out before. It's not going to revolutionise computer science. But it is an interesting observation.


    --
    Copyright 2022 Pete 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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  • From Richard Damon@21:1/5 to olcott on Thu Jun 2 08:18:45 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/22 8:01 AM, olcott wrote:
    On 6/2/2022 6:19 AM, Malcolm McLean wrote:
    On Wednesday, 1 June 2022 at 20:03:43 UTC+1, Ben wrote:
    Malcolm McLean <malcolm.ar...@gmail.com> writes:

    And actually, what you have done is construct a simulating attempted
    halt
    decider, which fails on H_Hat, for reasons unrelated to the invert
    loop.
    What? H_Hat(H_Hat) halts because H(H_Hat, H_Hat) == 0 so, yes, no loop
    is involved, but that's not interesting.
    So you have actually achieved something of interest, if you could only >>>> recognise that.
    What do you think is new or interesting here? I can't see anything new
    or interesting here at all.

    The "simulating halt decider" will fail when fed H_Hat, not because of
    the invert
    step, but because it can't solve the simulation of H. That's inherent
    in the way
    it is set up - it can only terminate if it detects that the series of
    nested
    simulations is non-terminating, in which case it becomes terminating.


    That is not the point. The simulated input to H(P,P) only halts if it
    reaches its "ret" instruction otherwise it is non-halting.

    No, the CORRECTLY simulated input will reach the "ret" instruction (if
    H(P,P) returns 0)

    All you have shown is that no H can be designed to reach the "ret"
    instruction in its own simumatlion of the input, but that if it aborts
    its simulation to avoid making the error of not answering, it make a
    mistake because it doesn't have enough information to make a correct answer.

    P(P) does halt yet it is an entirely different sequence of x86
    instructions than the correctly simulated input to H(P,P).


    No, it is EXACTLY the same sequence of x86 instuctons as the input to H.
    If it isn't, then your input is incorrect. Your problem is that the
    "sequence of x86 instructions" that represent the PROGRAM P, include all
    the instructions of the subroutine H (and everything it calls). You H
    acts as if those instructions were different then they actually are, so
    doesn't create a CORRECT simulation of that input.

    BAD LOGIC, BAD ANSWERS, DUMB logicitian.

    People here say that they really really don't believe this thus directly disagree with the x86 language. This is a move that only one person here
    is dumb enough to make, yet more than one person does.

    So, what do YOU think the x86 processor manual says the "call"
    instruction does?



    I don't think anyone has actually pointed that out before. It's not
    going to
    revolutionise computer science. But it is an interesting observation.



    Nope, you just prove how ignorant you are.

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  • From olcott@21:1/5 to Ben on Thu Jun 2 13:42:55 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 11:21 AM, Ben wrote:
    Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:

    On Wednesday, 1 June 2022 at 20:03:43 UTC+1, Ben wrote:
    Malcolm McLean <malcolm.ar...@gmail.com> writes:

    And actually, what you have done is construct a simulating attempted halt >>>> decider, which fails on H_Hat, for reasons unrelated to the invert
    loop.
    What? H_Hat(H_Hat) halts because H(H_Hat, H_Hat) == 0 so, yes, no loop
    is involved, but that's not interesting.
    So you have actually achieved something of interest, if you could only >>>> recognise that.
    What do you think is new or interesting here? I can't see anything new
    or interesting here at all.

    The "simulating halt decider" will fail when fed H_Hat, not because of
    the invert step, but because it can't solve the simulation of H.

    What's new or interesting about that? A simulating partial halt decider really only has only two ways to be wrong: fail to return a result or incorrectly decide some halting computations as non-halting (ignoring
    cases where it's designed to be deliberately wrong unnecessarily).

    And H_Hat never has anything to do with why a decider is wrong because
    we can show that any purported halt decider must get infinitely many instances wrong, so there will be an abundance of examples that have
    nothing to do with H_Hat.

    But none of this is what PO is claiming. He claims H is right to be
    wrong so you have your work cut out if you want to get him to accept
    that H really is wrong, even if you find the reason interesting.

    NO I AM NOT SAYING THAT AT ALL PLEASE PAY MUCH CLOSER ATTENTION I HAVE
    HAD TO REPEAT MYSELF HUNDREDS OF TIMES AND EVERYONE CONSISTENTLY IGNORES
    MY CORRECTIONS TO THEIR FALSE ASSUMPTIONS AND MISCONCEPTIONS.

    It is an easily verified fact that the correctly simulated input to
    H(P,P) never reaches its "ret" instruction therefore even when the
    simulation is aborted P did not halt and never would halt.

    This conclusively proves that H(P,P)==0 to anyone that is merely
    a fully competent software engineer having the required prerequisites
    of C, x86, how C translates to x86, and what x86 emulation is.

    LIARS AND IDIOTS DENY THE TRUTH OF THIS
    It is completely obvious that when H(P,P) correctly emulates its input
    that it must emulate the first seven instructions of P. Because the
    seventh instruction repeats this process we can know with complete
    certainty that the emulated P never reaches its find “ret” instruction, thus never halts.

    Because Ben very very persistently refused to confirm the truth of this
    I mistook him for a liar. When he acknowledged that he saw no evidence
    that it is not true I acknowledged that my attribution of liar was
    incorrect.



    That's inherent in the way
    it is set up - it can only terminate if it detects that the series of nested >> simulations is non-terminating, in which case it becomes terminating.

    No, it terminates because it /incorrectly/ concludes that the nested simulations are non-terminating. PO's silly argument is that this
    conclusion is not wrong because the nested simulations /would/ be non-terminating if they were not terminated. Go figure.

    I don't think anyone has actually pointed that out before. It's not going to >> revolutionise computer science. But it is an interesting observation.

    Why would anyone want to point this out? We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong. Both come up every time this
    is taught to a class of programmers. (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)



    --
    Copyright 2022 Pete 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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  • From olcott@21:1/5 to Mike Terry on Thu Jun 2 13:47:57 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 11:32 AM, Mike Terry wrote:
    On 02/06/2022 12:19, Malcolm McLean wrote:
    On Wednesday, 1 June 2022 at 20:03:43 UTC+1, Ben wrote:
    Malcolm McLean <malcolm.ar...@gmail.com> writes:

    And actually, what you have done is construct a simulating attempted
    halt
    decider, which fails on H_Hat, for reasons unrelated to the invert
    loop.
    What? H_Hat(H_Hat) halts because H(H_Hat, H_Hat) == 0 so, yes, no loop
    is involved, but that's not interesting.
    So you have actually achieved something of interest, if you could only >>>> recognise that.
    What do you think is new or interesting here? I can't see anything new
    or interesting here at all.

    The "simulating halt decider" will fail when fed H_Hat, not because of
    the invert
    step, but because it can't solve the simulation of H. That's inherent
    in the way
    it is set up - it can only terminate if it detects that the series of
    nested
    simulations is non-terminating, in which case it becomes terminating.

    ..in which case it becomes terminating BECAUSE OF THE INVERT STEP.


    AS I HAVE PROVEN MANY HUNDREDS OF TIMES THE SIMULATED INPUT TO H(P,P)
    NEVER REACHES THE INVERT STEP.

    It is an easily verified fact that the correctly simulated input to
    H(P,P) never reaches its "ret" instruction therefore even when the
    simulation is aborted P did not halt and never would halt.

    This conclusively proves that H(P,P)==0 to anyone that is merely
    a fully competent software engineer having the required prerequisites
    of C, x86, how C translates to x86, and what x86 emulation is.

    LIARS AND IDIOTS DENY THE TRUTH OF THIS
    It is completely obvious that when H(P,P) correctly emulates its input
    that it must emulate the first seven instructions of P. Because the
    seventh instruction repeats this process we can know with complete
    certainty that the emulated P never reaches its find “ret” instruction, thus never halts.


    --
    Copyright 2022 Pete 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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  • From olcott@21:1/5 to Andy Walker on Thu Jun 2 13:58:17 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 1:12 PM, Andy Walker wrote:
    On 02/06/2022 17:21, Ben wrote:
    [...]  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time this
    is taught to a class of programmers.  (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)

        My first thought was "no, it's the same", but on reflexion, and AFAIR,
    our mathematicians simply accepted what I told them, which is
    essentially what
    is at

        http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

        http://www.cuboid.me.uk/anw/G12FCO/lect18.html

    [both lectures from a module that I gave many moons ago].  We got pretty bright students, among the top four or five cohorts in the UK at that time [1990s] [but that was our high point, even beating Oxford one year], and of course many of them went into IT as a career.  They found NP-completeness much harder than UTMs, HP and related stuff.

        [The code G12FCO means G1 -- maths, 2 -- second year, FCO -- Formal Computation.]


    The HP where a decider decides what another decider will do with an
    input (Like Sipser) is too convoluted. I only address what a halt
    decider does with its input (Like Linz).

    Any fully competent software engineer can easily verify that H(P,P)==0
    is correct.

    #include <stdint.h>
    #define u32 uint32_t

    void P(u32 x)
    {
    if (H(x, x))
    HERE: goto HERE;
    return;
    }

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

    _P()
    [00001352](01) 55 push ebp
    [00001353](02) 8bec mov ebp,esp
    [00001355](03) 8b4508 mov eax,[ebp+08]
    [00001358](01) 50 push eax // push P
    [00001359](03) 8b4d08 mov ecx,[ebp+08]
    [0000135c](01) 51 push ecx // push P
    [0000135d](05) e840feffff call 000011a2 // call H
    [00001362](03) 83c408 add esp,+08
    [00001365](02) 85c0 test eax,eax
    [00001367](02) 7402 jz 0000136b
    [00001369](02) ebfe jmp 00001369
    [0000136b](01) 5d pop ebp
    [0000136c](01) c3 ret
    Size in bytes:(0027) [0000136c]

    It is completely obvious that when H(P,P) correctly emulates its input
    that it must emulate the first seven instructions of P. Because the
    seventh instruction repeats this process we can know with complete
    certainty that the emulated P never reaches its final “ret” instruction, thus never halts.

    Halting problem undecidability and infinitely nested simulation (V5) https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5


    --
    Copyright 2022 Pete 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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  • From olcott@21:1/5 to Ben on Thu Jun 2 14:54:02 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 2:00 PM, Ben wrote:
    Andy Walker <anw@cuboid.co.uk> writes:

    On 02/06/2022 17:21, Ben wrote:
    [...] We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong. Both come up every time this
    is taught to a class of programmers. (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)

    My first thought was "no, it's the same", but on reflexion, and AFAIR, >> our mathematicians simply accepted what I told them, which is essentially what
    is at

    http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

    http://www.cuboid.me.uk/anw/G12FCO/lect18.html

    I see your notes are an explicit counterexample to PO's claims that
    emulation is never considered!


    For these cases, we can turn to our second weapon -- emulation. We want
    to know whether a program halts, so we try it. If it halts, then we know
    the answer. If it doesn't halt, then `it must be in a loop', so we
    monitor its state and `detect the loop'. Sadly, although this is in one
    sense correct, it is a false dichotomy. At any given moment as the
    emulation proceeds, we are in one of not two but three states: the
    program has halted, or it is looping, or it is still running and has not
    yet entered a loop. It's the third case that kills us -- we just have to
    keep going, and wait for one of the other two things to happen. The
    trouble is that it may be that neither of them ever happens -- which is
    why `it must be in a loop' was in quotes above.

    It is not considered correctly.
    (a) the program has halted
    (b) the program is still running
    (c) the program matched an infinite behavior pattern

    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

    H(P,P)==0 for the above behavior pattern. (The entire research scope)


    [both lectures from a module that I gave many moons ago]. We got pretty
    bright students, among the top four or five cohorts in the UK at that time >> [1990s] [but that was our high point, even beating Oxford one year], and of >> course many of them went into IT as a career. They found NP-completeness
    much harder than UTMs, HP and related stuff.

    We got pretty bright students too. Bright students ask good questions
    but eventually (rapidly even) settle the matter to their own
    satisfaction.

    And I agree that complexity is more... complex. The details become so
    very significant in complexity theory. One of the depressing things
    about these endless threads is that this is not a complex theorem.



    --
    Copyright 2022 Pete 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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  • From olcott@21:1/5 to Ben on Thu Jun 2 14:42:41 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 2:00 PM, Ben wrote:
    Andy Walker <anw@cuboid.co.uk> writes:

    On 02/06/2022 17:21, Ben wrote:
    [...] We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong. Both come up every time this
    is taught to a class of programmers. (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)

    My first thought was "no, it's the same", but on reflexion, and AFAIR, >> our mathematicians simply accepted what I told them, which is essentially what
    is at

    http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

    http://www.cuboid.me.uk/anw/G12FCO/lect18.html

    I see your notes are an explicit counterexample to PO's claims that
    emulation is never considered!


    The emulation is simply describing a UTM not applying an adapted UTM to
    the HP.

    [both lectures from a module that I gave many moons ago]. We got pretty
    bright students, among the top four or five cohorts in the UK at that time >> [1990s] [but that was our high point, even beating Oxford one year], and of >> course many of them went into IT as a career. They found NP-completeness
    much harder than UTMs, HP and related stuff.

    We got pretty bright students too. Bright students ask good questions
    but eventually (rapidly even) settle the matter to their own
    satisfaction.

    And I agree that complexity is more... complex. The details become so
    very significant in complexity theory. One of the depressing things
    about these endless threads is that this is not a complex theorem.



    --
    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 olcott@21:1/5 to Andy Walker on Thu Jun 2 15:47:22 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 1:12 PM, Andy Walker wrote:
    On 02/06/2022 17:21, Ben wrote:
    [...]  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time this
    is taught to a class of programmers.  (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)

        My first thought was "no, it's the same", but on reflexion, and AFAIR,
    our mathematicians simply accepted what I told them, which is
    essentially what
    is at

        http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

        http://www.cuboid.me.uk/anw/G12FCO/lect18.html



    For these cases, we can turn to our second weapon -- emulation. We want
    to know whether a program halts, so we try it. If it halts, then we know
    the answer. If it doesn't halt, then `it must be in a loop', so we
    monitor its state and `detect the loop'. Sadly, although this is in one
    sense correct, it is a false dichotomy. At any given moment as the
    emulation proceeds, we are in one of not two but three states: the
    program has halted, or it is looping, or it is still running and has not
    yet entered a loop. It's the third case that kills us -- we just have to
    keep going, and wait for one of the other two things to happen. The
    trouble is that it may be that neither of them ever happens -- which is
    why `it must be in a loop' was in quotes above.

    It is not considered correctly.
    (a) the program has halted
    (b) the program is still running
    (c) the program matched an infinite behavior pattern

    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

    Any competent software engineer can verify that H(P,P)==0
    for the above behavior pattern. (The entire research scope)
    As detailed in my paper:

    Halting problem undecidability and infinitely nested simulation (V5)

    https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5


    [both lectures from a module that I gave many moons ago].  We got pretty bright students, among the top four or five cohorts in the UK at that time [1990s] [but that was our high point, even beating Oxford one year], and of course many of them went into IT as a career.  They found NP-completeness much harder than UTMs, HP and related stuff.

        [The code G12FCO means G1 -- maths, 2 -- second year, FCO -- Formal Computation.]



    --
    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 olcott@21:1/5 to Mr Flibble on Thu Jun 2 18:45:56 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 6:38 PM, Mr Flibble wrote:
    On Thu, 2 Jun 2022 15:47:22 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/2/2022 1:12 PM, Andy Walker wrote:
    On 02/06/2022 17:21, Ben wrote:
    [...]  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time
    this is taught to a class of programmers.  (I've never taught it
    to a class of mathematicians but I suspect the discussion would be
    very different.)

        My first thought was "no, it's the same", but on reflexion,
    and AFAIR, our mathematicians simply accepted what I told them,
    which is essentially what
    is at

        http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

        http://www.cuboid.me.uk/anw/G12FCO/lect18.html



    For these cases, we can turn to our second weapon -- emulation. We
    want to know whether a program halts, so we try it. If it halts, then
    we know the answer. If it doesn't halt, then `it must be in a loop',
    so we monitor its state and `detect the loop'. Sadly, although this
    is in one sense correct, it is a false dichotomy. At any given moment
    as the emulation proceeds, we are in one of not two but three states:
    the program has halted, or it is looping, or it is still running and
    has not yet entered a loop. It's the third case that kills us -- we
    just have to keep going, and wait for one of the other two things to
    happen. The trouble is that it may be that neither of them ever
    happens -- which is why `it must be in a loop' was in quotes above.

    It is not considered correctly.
    (a) the program has halted
    (b) the program is still running
    (c) the program matched an infinite behavior pattern

    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

    Any competent software engineer can verify that H(P,P)==0
    for the above behavior pattern. (The entire research scope)
    As detailed in my paper:

    Halting problem undecidability and infinitely nested simulation (V5)

    https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5


    The proofs you are attempting to refute do not contain the infinite
    behaviour pattern you describe;

    The correctly simulated input to a simulating halt decider never reaches
    the final instruction of this simulated input thus is unequivocally non-halting.

    Halting does not mean stopped running, it means terminated normally.

    Whether or not this simulated input specifies infinite behavior is a
    matter of how one defines one's terms and makes no difference because
    the behavior that it does exhibit is still correctly construed as
    non-halting even after its simulation has been aborted.

    you have been told this multiple times
    now and keep ignoring it:

    Every time that I correct my reviewers false assumptions and
    misconceptions they always make sure to not read a single word of what I
    have said.

    It is always blah blah blah we know you must be wrong so there is no
    sense in reading what your reply.

    is this because you have actually given up
    trying to refute those proofs?

    /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

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  • From Mr Flibble@21:1/5 to olcott on Fri Jun 3 00:38:06 2022
    XPost: comp.theory, sci.logic, sci.math

    On Thu, 2 Jun 2022 15:47:22 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/2/2022 1:12 PM, Andy Walker wrote:
    On 02/06/2022 17:21, Ben wrote:
    [...] We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong. Both come up every time
    this is taught to a class of programmers. (I've never taught it
    to a class of mathematicians but I suspect the discussion would be
    very different.)

    My first thought was "no, it's the same", but on reflexion,
    and AFAIR, our mathematicians simply accepted what I told them,
    which is essentially what
    is at

    http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

    http://www.cuboid.me.uk/anw/G12FCO/lect18.html



    For these cases, we can turn to our second weapon -- emulation. We
    want to know whether a program halts, so we try it. If it halts, then
    we know the answer. If it doesn't halt, then `it must be in a loop',
    so we monitor its state and `detect the loop'. Sadly, although this
    is in one sense correct, it is a false dichotomy. At any given moment
    as the emulation proceeds, we are in one of not two but three states:
    the program has halted, or it is looping, or it is still running and
    has not yet entered a loop. It's the third case that kills us -- we
    just have to keep going, and wait for one of the other two things to
    happen. The trouble is that it may be that neither of them ever
    happens -- which is why `it must be in a loop' was in quotes above.

    It is not considered correctly.
    (a) the program has halted
    (b) the program is still running
    (c) the program matched an infinite behavior pattern

    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

    Any competent software engineer can verify that H(P,P)==0
    for the above behavior pattern. (The entire research scope)
    As detailed in my paper:

    Halting problem undecidability and infinitely nested simulation (V5)

    https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5


    The proofs you are attempting to refute do not contain the infinite
    behaviour pattern you describe; you have been told this multiple times
    now and keep ignoring it: is this because you have actually given up
    trying to refute those proofs?

    /Flibble

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  • From Richard Damon@21:1/5 to olcott on Thu Jun 2 22:48:10 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/22 2:42 PM, olcott wrote:
    On 6/2/2022 11:21 AM, Ben wrote:
    Malcolm McLean <malcolm.arthur.mclean@gmail.com> writes:

    On Wednesday, 1 June 2022 at 20:03:43 UTC+1, Ben wrote:
    Malcolm McLean <malcolm.ar...@gmail.com> writes:

    And actually, what you have done is construct a simulating
    attempted halt
    decider, which fails on H_Hat, for reasons unrelated to the invert
    loop.
    What? H_Hat(H_Hat) halts because H(H_Hat, H_Hat) == 0 so, yes, no loop >>>> is involved, but that's not interesting.
    So you have actually achieved something of interest, if you could only >>>>> recognise that.
    What do you think is new or interesting here? I can't see anything new >>>> or interesting here at all.

    The "simulating halt decider" will fail when fed H_Hat, not because of
    the invert step, but because it can't solve the simulation of H.

    What's new or interesting about that?  A simulating partial halt decider
    really only has only two ways to be wrong: fail to return a result or
    incorrectly decide some halting computations as non-halting (ignoring
    cases where it's designed to be deliberately wrong unnecessarily).

    And H_Hat never has anything to do with why a decider is wrong because
    we can show that any purported halt decider must get infinitely many
    instances wrong, so there will be an abundance of examples that have
    nothing to do with H_Hat.

    But none of this is what PO is claiming.  He claims H is right to be
    wrong so you have your work cut out if you want to get him to accept
    that H really is wrong, even if you find the reason interesting.

    NO I AM NOT SAYING THAT AT ALL PLEASE PAY MUCH CLOSER ATTENTION I HAVE
    HAD TO REPEAT MYSELF HUNDREDS OF TIMES AND EVERYONE CONSISTENTLY IGNORES
    MY CORRECTIONS TO THEIR FALSE ASSUMPTIONS AND MISCONCEPTIONS.

    It is an easily verified fact that the correctly simulated input to
    H(P,P) never reaches its "ret" instruction therefore even when the
    simulation is aborted P did not halt and never would halt.

    Really, then what have traces of this input being correctly simulated
    been posted showing that it does reach the re tstatement.

    Perhaps the problem is that you are using wrong definitions of what a
    correct simulation is?

    If you mean something besides the simulation that exactly duplicates the behavor of the machine that the input rerpesents, you need to CLEARLY
    state what you think the definition is and show why it is correct.

    Note, just saying that you think it makes more sense isn't actually a
    good reason.


    This conclusively proves that H(P,P)==0 to anyone that is merely
    a fully competent software engineer having the required prerequisites
    of C, x86, how C translates to x86, and what x86 emulation is.


    No, it just shows that you don't know what the right definitions are.

    LIARS AND IDIOTS DENY THE TRUTH OF THIS
    It is completely obvious that when H(P,P) correctly emulates its input
    that it must emulate the first seven instructions of P. Because the
    seventh instruction repeats this process we can know with complete
    certainty that the emulated P never reaches its find “ret” instruction, thus never halts.

    OK, so you show your ignorance as to what a correct emulation of
    assembly code is.

    The is ZERO grounds for the "break" in your simulation, thus it is
    INCORRECT, and your instance that it is "correct" proves you to be a liar.

    The ONLY correct simulation of the input needs to show the exectution of
    the eighth instruction too, since that is what the code IS.

    You failure to see that shows that you just don't understand how
    computers works.


    Because Ben very very persistently refused to confirm the truth of this
    I mistook him for a liar. When he acknowledged that he saw no evidence
    that it is not true I acknowledged that my attribution of liar was
    incorrect.



    That's inherent in the way
    it is set up - it can only terminate if it detects that the series of
    nested
    simulations is non-terminating, in which case it becomes terminating.

    No, it terminates because it /incorrectly/ concludes that the nested
    simulations are non-terminating.  PO's silly argument is that this
    conclusion is not wrong because the nested simulations /would/ be
    non-terminating if they were not terminated.  Go figure.

    I don't think anyone has actually pointed that out before. It's not
    going to
    revolutionise computer science. But it is an interesting observation.

    Why would anyone want to point this out?  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time this
    is taught to a class of programmers.  (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)




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  • From Richard Damon@21:1/5 to olcott on Thu Jun 2 22:54:13 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/22 2:47 PM, olcott wrote:
    On 6/2/2022 11:32 AM, Mike Terry wrote:
    On 02/06/2022 12:19, Malcolm McLean wrote:
    On Wednesday, 1 June 2022 at 20:03:43 UTC+1, Ben wrote:
    Malcolm McLean <malcolm.ar...@gmail.com> writes:

    And actually, what you have done is construct a simulating
    attempted halt
    decider, which fails on H_Hat, for reasons unrelated to the invert
    loop.
    What? H_Hat(H_Hat) halts because H(H_Hat, H_Hat) == 0 so, yes, no loop >>>> is involved, but that's not interesting.
    So you have actually achieved something of interest, if you could only >>>>> recognise that.
    What do you think is new or interesting here? I can't see anything new >>>> or interesting here at all.

    The "simulating halt decider" will fail when fed H_Hat, not because
    of the invert
    step, but because it can't solve the simulation of H. That's inherent
    in the way
    it is set up - it can only terminate if it detects that the series of
    nested
    simulations is non-terminating, in which case it becomes terminating.

    ..in which case it becomes terminating BECAUSE OF THE INVERT STEP.


    AS I HAVE PROVEN MANY HUNDREDS OF TIMES THE SIMULATED INPUT TO H(P,P)
    NEVER REACHES THE INVERT STEP.

    You hav shown that the PARTIAL simulation by H never reaches that step.

    You CLAIM this is a correct simulation,



    It is an easily verified fact that the correctly simulated input to
    H(P,P) never reaches its "ret" instruction therefore even when the
    simulation is aborted P did not halt and never would halt.

    Nope, LIE. FAIL.


    This conclusively proves that H(P,P)==0 to anyone that is merely
    a fully competent software engineer having the required prerequisites
    of C, x86, how C translates to x86, and what x86 emulation is.

    Nope, LIE, FAIL.


    LIARS AND IDIOTS DENY THE TRUTH OF THIS
    It is completely obvious that when H(P,P) correctly emulates its input
    that it must emulate the first seven instructions of P. Because the
    seventh instruction repeats this process we can know with complete
    certainty that the emulated P never reaches its find “ret” instruction, thus never halts.



    YOU ARE THE LIAR because you use FALSE definitions.

    It is actually completely true that a real correct simulation of the
    input to HH(P,P) must match the behavior of P(P), and go through all the instructions of the copy of H that P calls, and see that H abort its
    simulation of ITS input, and returning its answer to the P that called
    it, and that P reaching its "ret" instruction.

    It is IMPOSSIBLE for H to do this, but that doesn't change the
    definition of a CORRECT simulation.

    The ONLY way that H can be a "Correct Simulator" is for it to NEVER
    abort its simulation, and then if fails to answer. Yes, if THAT is your
    H, then P(P) is non-halting, and H(P,P) returning zero WOULD have been a
    right answer for the problem, but NOT correct behavior for the machine,
    as we started with the premise that H didn't abort (or that it actual
    was a correct simulator, which is the same thing).

    You just show that you don't understand the meaning of the terms that
    you use.

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  • From Richard Damon@21:1/5 to olcott on Thu Jun 2 22:41:29 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/22 7:45 PM, olcott wrote:
    On 6/2/2022 6:38 PM, Mr Flibble wrote:
    On Thu, 2 Jun 2022 15:47:22 -0500
    olcott <NoOne@NoWhere.com> wrote:

    On 6/2/2022 1:12 PM, Andy Walker wrote:
    On 02/06/2022 17:21, Ben wrote:
    [...]  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time
    this is taught to a class of programmers.  (I've never taught it
    to a class of mathematicians but I suspect the discussion would be
    very different.)

          My first thought was "no, it's the same", but on reflexion, >>>> and AFAIR, our mathematicians simply accepted what I told them,
    which is essentially what
    is at

          http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

          http://www.cuboid.me.uk/anw/G12FCO/lect18.html


    For these cases, we can turn to our second weapon -- emulation. We
    want to know whether a program halts, so we try it. If it halts, then
    we know the answer. If it doesn't halt, then `it must be in a loop',
    so we monitor its state and `detect the loop'. Sadly, although this
    is in one sense correct, it is a false dichotomy. At any given moment
    as the emulation proceeds, we are in one of not two but three states:
    the program has halted, or it is looping, or it is still running and
    has not yet entered a loop. It's the third case that kills us -- we
    just have to keep going, and wait for one of the other two things to
    happen. The trouble is that it may be that neither of them ever
    happens -- which is why `it must be in a loop' was in quotes above.

    It is not considered correctly.
    (a) the program has halted
    (b) the program is still running
    (c) the program matched an infinite behavior pattern

           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

    Any competent software engineer can verify that H(P,P)==0
    for the above behavior pattern. (The entire research scope)
    As detailed in my paper:

    Halting problem undecidability and infinitely nested simulation (V5)

    https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5



    The proofs you are attempting to refute do not contain the infinite
    behaviour pattern you describe;

    The correctly simulated input to a simulating halt decider never reaches
    the final instruction of this simulated input thus is unequivocally non-halting.

    You keep on saying that, but it isn't actually true for the PROPER
    definition of "correct simulation", that being, the simulation of the
    input by a UTM equiavlent. (or an x86 processor for your x86 code).

    What you want to call "correct simulation" is the partial simulation by
    a particular H, but partial simulation are NEVER "correct" because they
    are, by definitin=on, INCOMPLETE.


    Halting does not mean stopped running, it means terminated normally., so we can

    Non-Halting also doesn't meen only partially run and it didn't reach
    completion yet.

    Note, the CORRECT simulation of the input to H will terminate normally,
    It is only that H didn't complete its simulation.

    Your arguement about "ANY H" is invalid, as H isn't just "ANY H" but is
    a particualar H, and each of the possible H's your "ANY H" talks about
    have DIFFERENT inputs so you can't use your "logic" to try to tight them together.

    That is a bit like saying the number N+5 is infinite for any finte
    number N, since no matter what N you choose, N+5 is bigger than you
    number, so that must mean that N+5 is bigger than all numbers, and thus infinite.

    Hopefully you can see the silliness of that claim, that is exact the
    sort of claim you are making with your "any H" claim.

    Given an H that aborts its simulation in "N" steps, we can show that
    P(P) will halt is some number of steps that can be computed as f(N) > N,
    but still finite. Your "ANY N" arguement to claim that P in non-halting
    is saying that f(N) is infinite, when it is actually a finite number
    just bigger then the N for that H.


    Whether or not this simulated input specifies infinite behavior is a
    matter of how one defines one's terms and makes no difference because
    the behavior that it does exhibit is still correctly construed as
    non-halting even after its simulation has been aborted.

    Nope. Since there is ONE definition of Halting, and that is does the
    ACTUAL MACHINE reach a final state in a finite number of steps, any
    defintion that can't be proved to be EXACTLY EQUIVALENT is BY DEFINITION incorrect.

    With your definition of "Correctly Simulated' being based on the
    simulation of the decider, that isn't true, so you can't use it as an
    alternate definition of halting, so your definition is just INCORRECT.


    you have been told this multiple times
    now and keep ignoring it:

    Every time that I correct my reviewers false assumptions and
    misconceptions they always make sure to not read a single word of what I
    have said.

    Nope, YOU don't read what people say and thus PROVE that you are just a pathological liar or a totally incompetent and ignorate fool.


    It is always blah blah blah we know you must be wrong so there is no
    sense in reading what your reply.

    So, what is actually wrong with what I said, or are YOU the one just
    going blah, blah, blah?

    (My guess is that is exactly what you will do, thus proving your status
    as a fool)


    is this because you have actually given up
    trying to refute those proofs?

    /Flibble




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  • From Richard Damon@21:1/5 to olcott on Thu Jun 2 23:07:04 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/22 3:54 PM, olcott wrote:
    On 6/2/2022 2:00 PM, Ben wrote:
    Andy Walker <anw@cuboid.co.uk> writes:

    On 02/06/2022 17:21, Ben wrote:
    [...]  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time this >>>> is taught to a class of programmers.  (I've never taught it to a class >>>> of mathematicians but I suspect the discussion would be very
    different.)

        My first thought was "no, it's the same", but on reflexion, and
    AFAIR,
    our mathematicians simply accepted what I told them, which is
    essentially what
    is at

        http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

        http://www.cuboid.me.uk/anw/G12FCO/lect18.html

    I see your notes are an explicit counterexample to PO's claims that
    emulation is never considered!


    For these cases, we can turn to our second weapon -- emulation. We want
    to know whether a program halts, so we try it. If it halts, then we know
    the answer. If it doesn't halt, then `it must be in a loop', so we
    monitor its state and `detect the loop'. Sadly, although this is in one
    sense correct, it is a false dichotomy. At any given moment as the
    emulation proceeds, we are in one of not two but three states: the
    program has halted, or it is looping, or it is still running and has not
    yet entered a loop. It's the third case that kills us -- we just have to
    keep going, and wait for one of the other two things to happen. The
    trouble is that it may be that neither of them ever happens -- which is
    why `it must be in a loop' was in quotes above.

    It is not considered correctly.
    (a) the program has halted
    (b) the program is still running
    (c) the program matched an infinite behavior pattern

         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

    H(P,P)==0 for the above behavior pattern. (The entire research scope)



    And your problem is that while H is simulating its input H^, that H^ is
    in the state (b) case, not state (c).

    This is shown because if H THINKS that H^ has entered state (c), then
    when we actually look that what H^ does, we see that after H has aborted
    it, that an ACTUAL CORRECT simulation of this input that does continue
    will see the H that this H^ is using also making that same (incorrect) decision, aborting its simulation, and returning to H^ and then H^
    entering state (a).

    There is no ACTUAL finite pattern that H sees in its simulation of H^
    applied to H^ that actually correctly matchs a correct "infinite behavor pattern".

    You just don't seem to be smart enough to understand this.


    [both lectures from a module that I gave many moons ago].  We got pretty >>> bright students, among the top four or five cohorts in the UK at that
    time
    [1990s] [but that was our high point, even beating Oxford one year],
    and of
    course many of them went into IT as a career.  They found
    NP-completeness
    much harder than UTMs, HP and related stuff.

    We got pretty bright students too.  Bright students ask good questions
    but eventually (rapidly even) settle the matter to their own
    satisfaction.

    And I agree that complexity is more... complex.  The details become so
    very significant in complexity theory.  One of the depressing things
    about these endless threads is that this is not a complex theorem.




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  • From olcott@21:1/5 to Andy Walker on Fri Jun 3 20:15:11 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/2/2022 1:12 PM, Andy Walker wrote:
    On 02/06/2022 17:21, Ben wrote:
    [...]  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time this
    is taught to a class of programmers.  (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)

        My first thought was "no, it's the same", but on reflexion, and AFAIR,
    our mathematicians simply accepted what I told them, which is
    essentially what
    is at

        http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

        http://www.cuboid.me.uk/anw/G12FCO/lect18.html

    [both lectures from a module that I gave many moons ago].  We got pretty bright students, among the top four or five cohorts in the UK at that time [1990s] [but that was our high point, even beating Oxford one year], and of course many of them went into IT as a career.  They found NP-completeness much harder than UTMs, HP and related stuff.

        [The code G12FCO means G1 -- maths, 2 -- second year, FCO -- Formal Computation.]


    At any given moment as the emulation proceeds, we are in one of not two
    but three states: the program has halted, or it is looping, or it is
    still running and has not yet entered a loop. It's the third case that
    kills us -- we just have to keep going, and wait for one of the other
    two things to happen. The trouble is that it may be that neither of them
    ever happens -- which is why `it must be in a loop' was in quotes above.

    Andy Walker did provide a fundamentally flawed and totally shallow
    analysis of an simulating halt decider.

    At any given moment as the emulation proceeds,
    we are in one of not two but three states:
    (a) The program has halted,
    (b) It is still running.
    (c) IT HAS MATCHED AN INFINITE BEHAVIOR PATTERN

    void P(u32 x)
    {
    if (H(x, x))
    HERE: goto HERE;
    return;
    }

    The above matches (c) for infinitely nested simulation.

    --
    Copyright 2022 Pete 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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  • From Richard Damon@21:1/5 to olcott on Fri Jun 3 22:44:34 2022
    XPost: comp.theory, sci.logic, sci.math

    On 6/3/22 9:15 PM, olcott wrote:
    On 6/2/2022 1:12 PM, Andy Walker wrote:
    On 02/06/2022 17:21, Ben wrote:
    [...]  We know simulation can't be
    used to decide halting, and we know that there are only two ways a
    decider that tries to do so can be wrong.  Both come up every time this >>> is taught to a class of programmers.  (I've never taught it to a class
    of mathematicians but I suspect the discussion would be very different.)

         My first thought was "no, it's the same", but on reflexion, and
    AFAIR,
    our mathematicians simply accepted what I told them, which is
    essentially what
    is at

         http://www.cuboid.me.uk/anw/G12FCO/lect17.html

    [second half, but see esp the last paragraph] and

         http://www.cuboid.me.uk/anw/G12FCO/lect18.html

    [both lectures from a module that I gave many moons ago].  We got pretty
    bright students, among the top four or five cohorts in the UK at that
    time
    [1990s] [but that was our high point, even beating Oxford one year],
    and of
    course many of them went into IT as a career.  They found NP-completeness >> much harder than UTMs, HP and related stuff.

         [The code G12FCO means G1 -- maths, 2 -- second year, FCO -- Formal
    Computation.]


    At any given moment as the emulation proceeds, we are in one of not two
    but three states: the program has halted, or it is looping, or it is
    still running and has not yet entered a loop. It's the third case that
    kills us -- we just have to keep going, and wait for one of the other
    two things to happen. The trouble is that it may be that neither of them
    ever happens -- which is why `it must be in a loop' was in quotes above.

    Andy Walker did provide a fundamentally flawed and totally shallow
    analysis of an simulating halt decider.

    At any given moment as the emulation proceeds,
    we are in one of not two but three states:
    (a) The program has halted,
    (b) It is still running.
    (c) IT HAS MATCHED AN INFINITE BEHAVIOR PATTERN

    void P(u32 x)
    {
      if (H(x, x))
        HERE: goto HERE;
      return;
    }

    The above matches (c) for infinitely nested simulation.


    No, it matches (B) unless H will NEVER abort its simulation. This is
    true because if H WILL abort its simulation, then this P will obviously
    Halt when H returns 0.

    Remember, H is a DEFINITE algorithm, so BY DEFINITION it WILL Either
    abort this input or it won't.

    If it does, then we never get to (c), because P will eventually get to (a).

    If it never get there, then yes P has entered (c), but H, because it
    never aborts its simulation, can never answer 0, so fails to be a decider.

    Only by LYING by claiming that H won't abort, to claim (c) and then
    aborting can you INCORRRECTLY, and by UNSOUND logic, claim that H was
    correct.

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