XPost: comp.theory, sci.logic, sci.math
On 6/23/2022 2:41 PM, Mr Flibble wrote:
On Thu, 23 Jun 2022 00:19:23 -0500
olcott <NoOne@NoWhere.com> wrote:
On 6/23/2022 12:13 AM, olcott wrote:
On 6/22/2022 10:41 PM, Richard Damon wrote:
On 6/22/22 10:55 PM, olcott wrote:
On 6/22/2022 9:34 PM, Richard Damon wrote:
On 6/22/22 10:18 PM, olcott wrote:
On 6/22/2022 8:45 PM, Richard Damon wrote:
On 6/22/22 9:41 PM, olcott wrote:
On 6/22/2022 8:36 PM, Richard Damon wrote:
On 6/22/22 9:29 PM, olcott wrote:
On 6/22/2022 8:14 PM, Richard Damon wrote:
On 6/22/22 8:55 PM, olcott wrote:
On 6/22/2022 7:48 PM, Richard Damon wrote:
On 6/22/22 8:37 PM, olcott wrote:
First you agree that my words are perfectly correct >>>>>>>>>>>>>>> within their specified context
Since you haven't actualy defined you context, and imply >>>>>>>>>>>>>> that it is the halting problem, where they can not be >>>>>>>>>>>>>> correct, that is not possible.
First you agree that these words are 100% correct within >>>>>>>>>>>>> the context of software engineering totally ignoring the >>>>>>>>>>>>> context of the halting problem.
#include <stdint.h>
#define u32 uint32_t
#include <stdint.h>
typedef void (*ptr)();
void P(ptr x)
{
  if (H(x, x))
    HERE: goto HERE;
  return;
}
int main()
{
  Output("Input_Halts = ", H(P, P));
}
_P()
[000010d2](01) 55             push ebp >>>>>>>>>>>>> [000010d3](02) 8bec           mov ebp,esp >>>>>>>>>>>>> [000010d5](03) 8b4508         mov eax,[ebp+08] >>>>>>>>>>>>> [000010d8](01) 50             push eax >>>>>>>>>>>>> [000010d9](03) 8b4d08         mov ecx,[ebp+08] >>>>>>>>>>>>> [000010dc](01) 51             push ecx >>>>>>>>>>>>> [000010dd](05) e820feffff     call 00000f02
[000010e2](03) 83c408         add esp,+08 >>>>>>>>>>>>> [000010e5](02) 85c0           test eax,eax >>>>>>>>>>>>> [000010e7](02) 7402           jz 000010eb >>>>>>>>>>>>> [000010e9](02) ebfe           jmp 000010e9 >>>>>>>>>>>>> [000010eb](01) 5d             pop ebp >>>>>>>>>>>>> [000010ec](01) c3             ret
Size in bytes:(0027) [000010ec]
Every sufficiently competent software engineer can easily >>>>>>>>>>>>> verify that the complete and correct x86 emulation of the >>>>>>>>>>>>> input to H(P,P) by H would never reach the "ret"
instruction of P because both H and P would remain stuck >>>>>>>>>>>>> in infinitely recursive emulation.
So, if H actually is a program that does a COMPLETE and >>>>>>>>>>>> correct x86 emulation of its input, then YES, as I have >>>>>>>>>>>> said many time before, this combination is non-halting. >>>>>>>>>>>>
The fact that you need to keep going back to this, and >>>>>>>>>>>> seem to just be refusing to accept the conditions under >>>>>>>>>>>> which you have proved it just shows the problems with your >>>>>>>>>>>> thought process.
If H does correctly determine that this is the case in a >>>>>>>>>>>>> finite number of steps then H could reject its input on >>>>>>>>>>>>> this basis. Here are the details of exactly how H does >>>>>>>>>>>>> this in a finite number of steps.
Except that NOW H isn't the H we were just talking about, >>>>>>>>>>>> so you are just proving that you are either lying or an >>>>>>>>>>>> idiot.
Remember, the first analysis had the CONDITION on it that >>>>>>>>>>>> H did a COMPLETE and correct x86 emulation.
Once you remove that property form H, that conclusion no >>>>>>>>>>>> long holds and you are shown to be a lying idiot.
typedef struct Decoded
{
  u32 Address;
  u32 ESP;         // Current value of ESP >>>>>>>>>>>>>   u32 TOS;         // Current value of Top of Stack >>>>>>>>>>>>>   u32 NumBytes;
  u32 Simplified_Opcode;
  u32 Decode_Target;
} Decoded_Line_Of_Code;
  machine  stack    stack    machine   assembly
  address  address  data     code      language
  ======== ======== ======== ========= ============= >>>>>>>>>>>>> [000010d2][00211e8a][00211e8e] 55        push ebp >>>>>>>>>>>>> [000010d3][00211e8a][00211e8e] 8bec      mov ebp,esp >>>>>>>>>>>>> [000010d5][00211e8a][00211e8e] 8b4508    mov eax,[ebp+08] >>>>>>>>>>>>> [000010d8][00211e86][000010d2] 50        push eax >>>>>>>>>>>>> // push P
[000010d9][00211e86][000010d2] 8b4d08Â Â Â Â mov ecx,[ebp+08] >>>>>>>>>>>>> [000010dc][00211e82][000010d2] 51Â Â Â Â Â Â Â Â push ecx >>>>>>>>>>>>> // push P
[000010dd][00211e7e][000010e2] e820feffff call 00000f02 >>>>>>>>>>>>> // call H
Infinitely Recursive Simulation Detected Simulation
Stopped
// actual fully operational code in the x86utm operating >>>>>>>>>>>>> system u32 H(u32 P, u32 I)
{
HERE:
  u32 End_Of_Code;
  u32 Address_of_H;             // 2022-06-17 >>>>>>>>>>>>>   u32 code_end                 = get_code_end(P);
  Decoded_Line_Of_Code *decoded =
(Decoded_Line_Of_Code*)
Allocate(sizeof(Decoded_Line_Of_Code)); Registers*
master_state     = (Registers*)
Allocate(sizeof(Registers)); Registers*Â slave_state >>>>>>>>>>>>> = (Registers*) Allocate(sizeof(Registers));
  u32*       slave_stack      = Allocate(0x10000); //
64k; u32Â execution_trace =
(u32)Allocate(sizeof(Decoded_Line_Of_Code) * 1000);
  __asm lea eax, HERE            // 2022-06-18 >>>>>>>>>>>>>   __asm sub eax, 6               // 2022-06-18
  __asm mov Address_of_H, eax    // 2022-06-18 >>>>>>>>>>>>>   __asm mov eax, END_OF_CODE
  __asm mov End_Of_Code, eax
  Output("Address_of_H:", Address_of_H); // 2022-06-11 >>>>>>>>>>>>>   Init_slave_state(P, I, End_Of_Code, slave_state, >>>>>>>>>>>>> slave_stack);
  Output("\nBegin Simulation  Execution Trace Stored >>>>>>>>>>>>> at:", execution_trace);
  if (Decide_Halting(&execution_trace, &decoded, >>>>>>>>>>>>> code_end, &master_state,
                     &slave_state, &slave_stack,
Address_of_H, P, I))
      goto END_OF_CODE;
  return 0; // Does not halt
END_OF_CODE:
  return 1; // Input has normally terminated
}
H knows its own machine address and on this basis it can >>>>>>>>>>>>> easily examine its stored execution_trace of P and
determine: (a) P is calling H with the same arguments >>>>>>>>>>>>> that H was called with.
(b) No instructions in P could possibly escape this
otherwise infinitely recursive emulation.
(c) H aborts its emulation of P before its call to H is >>>>>>>>>>>>> invoked.
(b) is NOT a correct rule. Thos has been pointed out
before, and you have ignored it.
That you don't understand what I mean does not mean that it >>>>>>>>>>> is an incorrect rule.
Here is an example where P does have instruction that could >>>>>>>>>>> possibly escape this otherwise infinitely recursive
emulation:
void P(ptr x)
{
static count = 0;
  count++;
  if count > 3)
    return;
  if (H(x, x))
    HERE: goto HERE;
  return;
}
FALLACY of proof by example. I never said that (b) isn't
sometimes true, just it isn't an always true condition. You >>>>>>>>>> fail at elementary logic.
Try and find a valid counter-example. Every attempt at
rebuttal that is not a valid counter-example is one form of
deception or another.
P(P)
That is not any example of (b), thus another mere strawman
deception.
Why not?
It is not an example of the simulation of the input to H(P,P) at
all.
Why is P not P?
It is not a direct rebuttal of my original claim.
This would be a direct rebuttal:
You must adapt P so that when H(P,P) emulates its input it
determines that P never reaches its "ret" instruction and the
adapted emulated P still reaches its "ret" instruction.
Unless you find that at least one input where it gets the wrong
answer you have not refuted me.
Here:
void Px(u32 x)
{
H(x, x);
return;
}
int main()
{
Output("Input_Halts = ", H((u32)Px, (u32)Px));
}
...[000013e8][00102357][00000000] 83c408 add esp,+08 ...[000013eb][00102353][00000000] 50 push eax ...[000013ec][0010234f][00000427] 6827040000 push 00000427 ---[000013f1][0010234f][00000427] e880f0ffff call 00000476
Input_Halts = 0
...[000013f6][00102357][00000000] 83c408 add esp,+08 ...[000013f9][00102357][00000000] 33c0 xor eax,eax ...[000013fb][0010235b][00100000] 5d pop ebp ...[000013fc][0010235f][00000004] c3 ret
Number of Instructions Executed(16120)
It gets the answer wrong, i.e. input has not been decided correctly.
QED.
/Flibble
To fully understand this code a software engineer must be an expert in:
the C programming language, the x86 programming language, exactly how C translates into x86 and the ability to recognize infinite recursion at
the x86 assembly language level. No knowledge of the halting problem is required.
The ordinary semantics of standard C and the conventional x86 language
are the entire semantics required to conclusively prove that H(Px,Px)
does correctly determine that its correct and complete x86 emulation of
its input would never reach the "ret" instruction of Px.
It is always the case that whenever anyone disagrees with verifiable
facts (as you have just done) that they are necessarily always incorrect.
--
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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