On Saturday, 23 April 2022 at 19:41:09 UTC+1, olcott wrote:
> On 4/23/2022 1:35 PM, Malcolm McLean wrote:
> > On Saturday, 23 April 2022 at 18:56:56 UTC+1, olcott wrote:
> >> On 4/23/2022 12:42 PM, wij wrote:
> >>> On Sunday, 24 April 2022 at 01:35:47 UTC+8, olcott wrote:
> >>>> On 4/23/2022 12:29 PM, wij wrote:
> >>>>> On Sunday, 24 April 2022 at 01:20:23 UTC+8, olcott wrote:
> >>>>>> On 4/23/2022 12:15 PM, wij wrote:
> >>>>>>> On Sunday, 24 April 2022 at 00:52:00 UTC+8, olcott wrote:
> >>>>>>>> On 4/23/2022 11:43 AM, wij wrote:
> >>>>>>>>> On Sunday, 24 April 2022 at 00:34:55 UTC+8, olcott wrote:
> >>>>>>>>>> On 4/23/2022 11:29 AM, wij wrote:
> >>>>>>>>>>> On Sunday, 24 April 2022 at 00:05:19 UTC+8, olcott wrote:
> >>>>>>>>>>>> On 4/23/2022 10:58 AM, wij wrote:
> >>>>>>>>>>>>> On Saturday, 23 April 2022 at 22:49:59 UTC+8, olcott wrote:
> >>>>>>>>>>>>>> All of my reviewers expect H(P,P) to compute the halt status of P(P),
> >>>>>>>>>>>>>> yet the behavior specified by the input to H(P,P) is not the same as the
> >>>>>>>>>>>>>> behavior specified by P(P).
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program and an input, whether the program will finish running, or continue to run forever
> >>>>>>>>>>>>> https://en.wikipedia.org/wiki/Halting_problem
> >>>>>>>>>>>>>
> >>>>>>>>>>>>> If using this common concept, the halting decider H, when given an argument P
> >>>>>>>>>>>>> (or P P), is supposed to answer whether P(P) will halt or not. This is a very
> >>>>>>>>>>>>> simple, easy idea to understand even for teenager students.
> >>>>>>>>>>>>>
> >>>>>>>>>>>> When this very simple idea is very rigorously examined (as it is in my
> >>>>>>>>>>>> paper) one sees that this requires the halt decider to be a mind reader
> >>>>>>>>>>>> and compute the halt status other than the actual halt status specified
> >>>>>>>>>>>> by its actual input.
> >>>>>>>>>>>
> >>>>>>>>>>> Your wording/interpretations/paper change all the time. No idea what this new
> >>>>>>>>>>> excuse 'mind reader' might mean. As said, the Halting Problem is very simple
> >>>>>>>>>>> and intuitive.
> >>>>>>>>>>> H should be a decider that computes the actual halt status of P(P). P is the
> >>>>>>>>>>> H's actual argument input.
> >>>>>>>>>>> I expect you might try to find some bugs of those descriptions to rephrasing it
> >>>>>>>>>>> in your favor. But, what would be the point? What is the usefulness of POOP?
> >>>>>>>>>> Yet when you carefully examine my paper:
> >>>>>>>>>> Anyone that is an expert in the C programming language, the x86
> >>>>>>>>>> programming language, exactly how C translates into x86 and what an x86
> >>>>>>>>>> processor emulator is can easily verify that the correctly simulated
> >>>>>>>>>> input to H(P,P) by H specifies a non-halting sequence of configurations.
> >>>>>>>>>> Halting problem undecidability and infinitely nested simulation (V5)
> >>>>>>>>>>
> >>>>>>>>>> https://www.researchgate.net/publication/359984584_Halting_problem_undecidability_and_infinitely_nested_simulation_V5
> >>>>>>>>>> Simply ignoring the verified facts is a ridiculously foolish was to form
> >>>>>>>>>> any actual rebuttal.
> >>>>>>>>>> --
> >>>>>>>>>> Copyright 2022 Pete Olcott
> >>>>>>>>>>
> >>>>>>>>>> "Talent hits a target no one else can hit;
> >>>>>>>>>> Genius hits a target no one else can see."
> >>>>>>>>>> Arthur Schopenhauer
> >>>>>>>>>
> >>>>>>>>> You already verified the fact that H(P,P) will be in an infinite recursive call
> >>>>>>>>> (thus, undecidable). Why you say H(P,P)==false (or true)?
> >>>>>>>> You might make a wild guess like this if you make sure to hardly pay
> >>>>>>>> attention. When you actually pay close attention and carefully study my
> >>>>>>>> paper it is very easy to see that H sees the same infinitely repeating
> >>>>>>>> pattern that we see, thus can abort its simulation and reject its input.
> >>>>>>>> Begin Local Halt Decider Simulation
> >>>>>>>>
> >>>>>>>> machine stack stack machine assembly
> >>>>>>>> address address data code language
> >>>>>>>> ======== ======== ======== ========= =============
> >>>>>>>> ...[000009d6][00211368][0021136c] 55 push ebp // enter P
> >>>>>>>> ...[000009d7][00211368][0021136c] 8bec mov ebp,esp
> >>>>>>>> ...[000009d9][00211368][0021136c] 8b4508 mov eax,[ebp+08]
> >>>>>>>> ...[000009dc][00211364][000009d6] 50 push eax // Push P
> >>>>>>>> ...[000009dd][00211364][000009d6] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>> ...[000009e0][00211360][000009d6] 51 push ecx // Push P
> >>>>>>>> ...[000009e1][0021135c][000009e6] e840feffff call 00000826 // Call H
> >>>>>>>> ...[000009d6][0025bd90][0025bd94] 55 push ebp // enter P
> >>>>>>>> ...[000009d7][0025bd90][0025bd94] 8bec mov ebp,esp
> >>>>>>>> ...[000009d9][0025bd90][0025bd94] 8b4508 mov eax,[ebp+08]
> >>>>>>>> ...[000009dc][0025bd8c][000009d6] 50 push eax // Push P
> >>>>>>>> ...[000009dd][0025bd8c][000009d6] 8b4d08 mov ecx,[ebp+08]
> >>>>>>>> ...[000009e0][0025bd88][000009d6] 51 push ecx // Push P
> >>>>>>>> ...[000009e1][0025bd84][000009e6] e840feffff call 00000826 // Call H
> >>>>>>>> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
> >>>>>>>> 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
> >>>>>>>
> >>>>>>> "Local Halt Decider: Infinite Recursion Detected Simulation Stopped" means
> >>>>>>> your x86utm emulator has encountered an infinite recursive call.
> >>>>>>> This is referred to as "undecidable". This is the fact.
> >>>>>> So you are saying that after H makes the correct halt status decision
> >>>>>> that this correct halt status decision is impossible to make.
> >>>>>>
> >>>>>> That is just like my example a smashing a Boston cream pie in your face
> >>>>>> and while this pie drips from your face you deny that the pie exists.
> >>>>>
> >>>>> Your H did not show 'correct decision' but 'unreachable' (exactly what the HP says).
> >>>>>
> >>>>> This is like "0.999..." (or repeating decimal) problems: Infinite repeating
> >>>>> simply means INFINITE repeating. Please, respect what it is.
> >>>> H simulates its input one x86 instruction at a time using an x86
> >>>> emulator. As soon as H(P,P) detects the same infinitely repeating
> >>>> pattern (that we can all see), it aborts its simulation and rejects its
> >>>> input.
> >>>> --
> >>>> Copyright 2022 Pete Olcott
> >>>>
> >>>> "Talent hits a target no one else can hit;
> >>>> Genius hits a target no one else can see."
> >>>> Arthur Schopenhauer
> >>>
> >>> Your H shows 'undecidable'. There is no branching codes shown.
> >> Without seeing any of the details of the 264 pages of the execution
> >> trace of H we can verify on the basis of the x86 source code for P and
> >> the execution trace that H derives that H does correctly simulate its
> >> input until it detects the same infinitely repeating pattern that we all
> >> can see.
> >>
> > Ben believes that H is just a call to P(P), and that there are no nested
> > simulations (simulations of simulations),
> Since I explained "how" I made my simulator re-entrant:
> Each simulation has its own registers, RAM and stack,
> and this is what is required to make my simulator re-entrant and every
> post that I have provided for many months shows that each subsequent
> simulation does have its own 64K stack your guess has been refuted.
> Begin Local Halt Decider Simulation
>
> machine stack stack machine assembly
> address address data code language
> ======== ======== ======== ========= =============
> ...[000009d6][00211368][0021136c] 55 push ebp // enter P
> ...[000009d7][00211368][0021136c] 8bec mov ebp,esp
> ...[000009d9][00211368][0021136c] 8b4508 mov eax,[ebp+08]
> ...[000009dc][00211364][000009d6] 50 push eax // Push P
> ...[000009dd][00211364][000009d6] 8b4d08 mov ecx,[ebp+08]
> ...[000009e0][00211360][000009d6] 51 push ecx // Push P
> ...[000009e1][0021135c][000009e6] e840feffff call 00000826 // Call H
> ...[000009d6][0025bd90][0025bd94] 55 push ebp // enter P
> ...[000009d7][0025bd90][0025bd94] 8bec mov ebp,esp
> ...[000009d9][0025bd90][0025bd94] 8b4508 mov eax,[ebp+08]
> ...[000009dc][0025bd8c][000009d6] 50 push eax // Push P
> ...[000009dd][0025bd8c][000009d6] 8b4d08 mov ecx,[ebp+08]
> ...[000009e0][0025bd88][000009d6] 51 push ecx // Push P
> ...[000009e1][0025bd84][000009e6] e840feffff call 00000826 // Call H
> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
> > and that the infinite cycle
> > detection code is in the outer layer, not incorporated into H at all.
> >
> Yes, Ben determines that I must be wrong entirely on the basis that he
> really believes that I am wrong. When we get down to actual verified
> facts, Ben has nothing and I have everything.
> > That would explain why the execution trace shows simuated instructions,
> > not the trace of the simulated simulator. it would also explain why you
> > are withholding H.
> >
> The outer-most H sees the infinite pattern first because it has seen
> more simulated steps than any of the inner simulations.
> > I'm reserving judgement. But it's impossible to verify whether a program
> > halts or not if portions of it are concealed.
> Anyone that is an expert in the C programming language, the x86
> programming language, exactly how C translates into x86 and what an x86
> processor emulator is can easily verify that the correctly simulated
> input to H(P,P) by H specifies a non-halting sequence of configurations.
>
Anyone who knows what an x86 emulator is knows that the execution trace
of the emulator looks nothing like the execution trace of the program it is
emulating. And anyone who knows even a small amount about x86 code
knows that you can't tell whether a call will return or not except by examining
the called code.
Then I'm very reluctant to accuse anyone of lying. But you could instantly show
that Ben is in the wrong to make this accusation by providing the source of H.
Since it is necessary to show H to make your case, since it would be very easy to
do so, and since it would force Ben to issue an apology, surely you must see
that people will conclude that Ben is in fact right?