On 6/13/2022 3:46 PM, Juha Nieminen wrote:
The very weird problem in this case is that software engineering has
caught a very subtle mistake that computer science textbooks have been
making regarding the halting problem. My paper details this mistake in
Appendix I.
Without getting into the details that are outside of the scope of this
forum all that I really need here is a step-by-step confirmation that
H(P,P) correctly determines that the correct x86 emulation of its input
would never reach the "ret" instruction of this input.
*I know that it is already as obvious as Hell*
#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
// H emulates the first seven instructions of P
...[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
// The emulated H emulates the first seven instructions of P
...[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
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 of P repeats this process we know with complete
certainty that the correct and complete emulation of P by H would never
reach the final “ret” instruction of P, thus never halts.
...[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 Instructions Executed(15892) = 237 pages