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computers / comp.ai.philosophy / Re: Halting problem undecidability and infinitely nested simulation (V5) (update)

Re: Halting problem undecidability and infinitely nested simulation (V5) (update)

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Subject: Re: Halting problem undecidability and infinitely nested simulation
(V5) (update)
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From: NoO...@NoWhere.com (olcott)
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 by: olcott - Sat, 21 May 2022 21:50 UTC

On 5/21/2022 4:48 PM, olcott wrote:
> Halting problem undecidability and infinitely nested simulation (V5)
>
> This is an explanation of a key new insight into the halting problem
> provided in the language of software engineering. Technical computer
> science terms are explained using software engineering terms.
>
> To fully understand this paper a software engineer must be an expert in:
> the C programming language, the x86 programming language, exactly how C
> translates into x86 and what an x86 processor emulator is. No knowledge
> of the halting problem is required.
>
> The computer science term “halting” means that a Turing Machine
> terminated normally reaching its last instruction known as its “final
> state”. This is the same idea as when a function returns to its caller
> as opposed to and contrast with getting stuck in an infinite loop or
> infinite recursion.
>
>      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. Alan
> Turing proved
>      in 1936 that a general algorithm to solve the halting problem for
> all possible
>      program-input pairs cannot exist.
>
>      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
>
> Technically a halt decider is a program that computes the mapping from a
> pair of input finite strings to its own accept or reject state based on
> the actual behavior specified by these finite strings.  In other words
> it determines whether or not its input would halt and returns 0 or 1
> accordingly.
>
>      Computable functions are the basic objects of study in
> computability theory.
>      Computable functions are the formalized analogue of the intuitive
> notion of
>      algorithms, in the sense that a function is computable if there
> exists an algorithm
>      that can do the job of the function, i.e. given an input of the
> function domain it
>      can return the corresponding output.
>      https://en.wikipedia.org/wiki/Computable_function
>
> The most definitive way to determine the actual behavior of the actual
> input is to simply simulate this input and watch its behavior. This is
> the ultimate measure of the actual behavior of the input. A simulating
> halt decider (SHD) simulates its input and determines the halt status of
> this input on the basis of the behavior of this correctly simulated of
> its input.
>
> The x86utm operating system was created so that all of the details of
> the the halting problem counter-example could be examined at the much
> higher level of abstraction of the C/x86 computer languages. It is based
> on a very powerful x86 emulator.
> The function named P was defined to do the opposite of whatever H
> reports that it will do. If H(P,P) reports that its input halts, P
> invokes an infinite loop. If H(P,P) reports that its input is
> non-halting, P immediately halts.
>
> The technical computer science term "halt" means that a program will
> reach its last instruction technically called its final state. For P
> this would be its machine address [0000136c].
>
> 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.
>
> 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.
>
> Software engineering experts can reverse-engineer what the correct x86
> emulation of the input to H(P,P) would be for one emulation and one
> nested emulation thus confirming that the provided execution trace is
> correct. They can do this entirely on the basis of the x86 source-code
> for P with no need to see the source-code or execution trace of H.
>
> The function named H continues to simulate its input using an x86
> emulator until this input either halts on its own or H detects that it
> would never halt. If its input halts H returns 1. If H detects that its
> input would never halt H returns 0.
>
> #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]
>
> _main()
> [00001372](01)  55              push ebp
> [00001373](02)  8bec            mov ebp,esp
> [00001375](05)  6852130000      push 00001352 // push P
> [0000137a](05)  6852130000      push 00001352 // push P
> [0000137f](05)  e81efeffff      call 000011a2 // call H
> [00001384](03)  83c408          add esp,+08
> [00001387](01)  50              push eax
> [00001388](05)  6823040000      push 00000423 // "Input_Halts = "
> [0000138d](05)  e8e0f0ffff      call 00000472 // call Output
> [00001392](03)  83c408          add esp,+08
> [00001395](02)  33c0            xor eax,eax
> [00001397](01)  5d              pop ebp
> [00001398](01)  c3              ret
> Size in bytes:(0039) [00001398]
>
>     machine   stack     stack     machine    assembly
>     address   address   data      code       language
>     ========  ========  ========  =========  =============
> ...[00001372][0010229e][00000000] 55         push ebp
> ...[00001373][0010229e][00000000] 8bec       mov ebp,esp
> ...[00001375][0010229a][00001352] 6852130000 push 00001352 // push P
> ...[0000137a][00102296][00001352] 6852130000 push 00001352 // push P
> ...[0000137f][00102292][00001384] e81efeffff call 000011a2 // call H
>
> Begin Local Halt Decider Simulation   Execution Trace Stored at:212352
> ...[00001352][0021233e][00212342] 55         push ebp      // enter P
> ...[00001353][0021233e][00212342] 8bec       mov ebp,esp
> ...[00001355][0021233e][00212342] 8b4508     mov eax,[ebp+08]
> ...[00001358][0021233a][00001352] 50         push eax      // push P
> ...[00001359][0021233a][00001352] 8b4d08     mov ecx,[ebp+08]
> ...[0000135c][00212336][00001352] 51         push ecx      // push P
> ...[0000135d][00212332][00001362] e840feffff call 000011a2 // call H
> ...[00001352][0025cd66][0025cd6a] 55         push ebp      // enter P
> ...[00001353][0025cd66][0025cd6a] 8bec       mov ebp,esp
> ...[00001355][0025cd66][0025cd6a] 8b4508     mov eax,[ebp+08]
> ...[00001358][0025cd62][00001352] 50         push eax      // push P
> ...[00001359][0025cd62][00001352] 8b4d08     mov ecx,[ebp+08]
> ...[0000135c][0025cd5e][00001352] 51         push ecx      // push P
> ...[0000135d][0025cd5a][00001362] e840feffff call 000011a2 // call H
> Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>
> H sees that P is calling the same function from the same machine address
> with identical parameters, twice in sequence. This is the infinite
> recursion (infinitely nested simulation) non-halting behavior pattern.
>
> ...[00001384][0010229e][00000000] 83c408     add esp,+08
> ...[00001387][0010229a][00000000] 50         push eax
> ...[00001388][00102296][00000423] 6823040000 push 00000423 //
> "Input_Halts = "
> ---[0000138d][00102296][00000423] e8e0f0ffff call 00000472 // call Output
> Input_Halts = 0
> ...[00001392][0010229e][00000000] 83c408     add esp,+08
> ...[00001395][0010229e][00000000] 33c0       xor eax,eax
> ...[00001397][001022a2][00100000] 5d         pop ebp
> ...[00001398][001022a6][00000004] c3         ret
> Number_of_User_Instructions(1)
> Number of Instructions Executed(15892) = 237 pages
>
> The correct simulation of the input to H(P,P) and the direct execution
> of P(P) are not computationally equivalent thus need not have the same
> halting behavior.
>

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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o Halting problem undecidability and infinitely nested simulation (V5)

By: olcott on Sat, 21 May 2022

20olcott
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