Re: Simulating termination analyzers for dummies

On 6/19/2024 3:08 AM, Fred. Zwarts wrote:

Op 18.jun.2024 om 18:26 schreef olcott:

On 6/18/2024 10:47 AM, Fred. Zwarts wrote:

Op 18.jun.2024 om 17:33 schreef olcott:

On 6/18/2024 10:20 AM, Fred. Zwarts wrote:
>
It is a verified fact that serious C people have recently
agreed to the following verbatim statement in the C group.

>
http://al.howardknight.net/?STYPE=msgid&MSGI=%3Cv4pg5p%24morv%241%40raubtier-asyl.eternal-september.org%3E+
>

You either lack this degree of skill in C or are only
interested in playing head games.

>
I have seen the response. It was most certainly not a serious reply.
But you know apparently to little of C to understand that.
Probably, because you are unable to escape from rebuttal mode, even if the truth is obvious.
>

>
I have known C since K&R was the standard and met
Bjarne Stroustrup when he came to our university
to promote his new C++ programming language.
>
*You seem to be willfully ignorant*
>

It was your own proof that showed that in
>
        int main()
        {
          return H(main);
        }
>
>
main halts, whereas H reported non-halting. So, it you were honest you would stop claiming that H is correct.
>

>
That is merely a more difficult to understand version of this
same pathological relationship.
>
int main()
{
   Output("Input_Halts = ", HH0(main));
}
>
_main()
[000020c2] 55         push ebp
[000020c3] 8bec       mov ebp,esp
[000020c5] 68c2200000 push 000020c2 ; push main
[000020ca] e833f4ffff call 00001502 ; call HH0
[000020cf] 83c404     add esp,+04
[000020d2] 50         push eax
[000020d3] 6843070000 push 00000743
[000020d8] e885e6ffff call 00000762
[000020dd] 83c408     add esp,+08
[000020e0] eb04       jmp 000020e6
[000020e2] 33c0       xor eax,eax
[000020e4] eb02       jmp 000020e8
[000020e6] 33c0       xor eax,eax
[000020e8] 5d         pop ebp
[000020e9] c3         ret
Size in bytes:(0040) [000020e9]
>
  machine   stack     stack     machine    assembly
  address   address   data      code       language
  ========  ========  ========  =========  =============
[000020c2][001036c3][00000000] 55         push ebp
[000020c3][001036c3][00000000] 8bec       mov ebp,esp
[000020c5][001036bf][000020c2] 68c2200000 push 000020c2 ; push main
[000020ca][001036bb][000020cf] e833f4ffff call 00001502 ; call HH0
New slave_stack at:103767
>
Begin Local Halt Decider Simulation   Execution Trace Stored at:11376f
[000020c2][0011375f][00113763] 55         push ebp      ; begin main
[000020c3][0011375f][00113763] 8bec       mov ebp,esp
[000020c5][0011375b][000020c2] 68c2200000 push 000020c2 ; push main
[000020ca][00113757][000020cf] e833f4ffff call 00001502 ; call HH0
New slave_stack at:14e18f
[000020c2][0015e187][0015e18b] 55         push ebp      ; begin main
[000020c3][0015e187][0015e18b] 8bec       mov ebp,esp
[000020c5][0015e183][000020c2] 68c2200000 push 000020c2 ; push main
[000020ca][0015e17f][000020cf] e833f4ffff call 00001502 ; call HH0
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
>
[000020cf][001036c3][00000000] 83c404     add esp,+04
[000020d2][001036bf][00000000] 50         push eax
[000020d3][001036bb][00000743] 6843070000 push 00000743
[000020d8][001036bb][00000743] e885e6ffff call 00000762
Input_Halts = 0
[000020dd][001036c3][00000000] 83c408     add esp,+08
[000020e0][001036c3][00000000] eb04       jmp 000020e6
[000020e6][001036c3][00000000] 33c0       xor eax,eax
[000020e8][001036c7][00000018] 5d         pop ebp
[000020e9][001036cb][00000000] c3         ret           ; exit main
Number of Instructions Executed(10070) == 150 Pages
>

 It is easier to understand because a print statement was added.
You proved that it halts, but H0 reports non-halting.
So, it produces a false negative.
So, now it has been proved that H, H0, etc produce false negatives, when used to determine halting behaviour, please, stop to call them halt-deciders, or termination-deciders.
They might be "simulation deciders". When returning true, the simulation was correct, when false, the full simulation was not possible.

I don't want to discuss your screwy example because I
can't use screwy examples in my paper.
void DDD()
{
   H0(DDD);
}
_DDD()
[000020a2] 55         push ebp      ; housekeeping
[000020a3] 8bec       mov ebp,esp   ; housekeeping
[000020a5] 68a2200000 push 000020a2 ; push DDD
[000020aa] e8f3f9ffff call 00001aa2 ; call H0
[000020af] 83c404     add esp,+04   ; housekeeping
[000020b2] 5d         pop ebp       ; housekeeping
[000020b3] c3         ret           ; never gets here
Size in bytes:(0018) [000020b3]
Exactly which step of DDD emulated by H0 was emulated
incorrectly such that this emulation would be complete?
AKA DDD emulated by H0 reaches machine address [000020b3]
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer