Re: People are still trying to get away with disagreeing with the semantics of the x86 language

On 6/29/2024 1:38 PM, Richard Damon wrote:

On 6/29/24 2:06 PM, olcott wrote:

On 6/29/2024 12:59 PM, Richard Damon wrote:

On 6/29/24 1:17 PM, olcott wrote:

On 6/29/2024 11:45 AM, Richard Damon wrote:

On 6/29/24 12:09 PM, olcott wrote:

People are still trying to get away with disagreeing with
the semantics of the x86 language. That is isomorphic to
trying to get away with disagreeing with arithmetic.

>
Nope, we are not disagreeing with the semantics of the x86 language, we are disagreeing with your misunderstanding of how it works.
>

>
typedef void (*ptr)();
int H0(ptr P);
>
void Infinite_Loop()
{
   HERE: goto HERE;
}
>
void Infinite_Recursion()
{
   Infinite_Recursion();
}
>
void DDD()
{
   H0(DDD);
}
>
int main()
{
   H0(Infinite_Loop);
   H0(Infinite_Recursion);
   H0(DDD);
}
>
Every C programmer that knows what an x86 emulator is knows
that when H0 emulates the machine language of Infinite_Loop,
Infinite_Recursion, and DDD that it must abort these emulations
so that itself can terminate normally.

>
No the x86 language "knows" NOTHING about H0 being a x86 emulator. It is just a function that maybe happens to be a partial x86 emulator, but that is NOT a fundamental result of it being H0.
>

>
When this is construed as non-halting criteria then simulating
termination analyzer H0 is correct to reject these inputs as
non-halting by returning 0 to its caller.

>
It is construed as non-halting BECAUSE it has been shown that your H0 *WILL* terminate its PARTIAL emulation of the code it is emulating and returning.
>

>
Simulating termination analyzers must report on the behavior
that their finite string input specifies thus H0 must report
that DDD correctly emulated by H0 remains stuck in recursive
simulation.

>
Right, so H0 is REQUIRED to return, and thus if the termination analyser knows that H0 is a termination analyzer it knows that the call to H0 MUST return, and thus DDD must be a terminating program.
>
An H0 that doesn't know this, and can't figure out that H0 will return, but just keeps emulating H0 emulating its input will just fail to meet its own requirement to return.
>

>
<MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
     If simulating halt decider H correctly simulates its input D
     until H correctly determines that its simulated D would never
     stop running unless aborted then
>
     H can abort its simulation of D and correctly report that D
     specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

>
Right, and the only definition Professor Sipser uses for "Correct Simulation" is a simulation that EXACTLY REPRODUCES the behavior of the directly executed program represented by the input. Your H doesn't do that, nor correctly predicts the behavior of such a simulation of the input (since that behavior is to halt) so it can never proper avail itself of the second paragraph, so does so erroneously getting the wrong answer.
>

>
People are trying to get away with disagreeing with the semantics
of the x86 language by disagreeing that
>
The call from DDD to HHH(DDD) when N steps of DDD are correctly
emulated by any pure function x86 emulator HHH cannot possibly
return.

>
Except that the "N Steps of DDD correctly emulated" is NOT the definition of the "behavior" of the input DDD.
>
"inputs" Do not have "behavoir", that is a property of a program, so the input only "represents" that program, in this case the program DDD.
>

>
*According to the professor Sipser approved criteria YES IT IS*
>

>
Nope. Where dp you see that in what he says? Remember, you need to interpret the words by what he means them to say.
>
His ONLY definition of "Correct Simulation" is a simulation that exactly recreates the behavior of the program described by the input, and that in one that does not stop its simulation. So, NOT your "N Step"
>

>
*N steps of correct simulation are specified*
H correctly simulates its input D until H
H correctly simulates its input D until H
H correctly simulates its input D until H
H correctly simulates its input D until H

 Which does not determine the ACTUAL behavor
 

_DDD()
[00002172] 55               push ebp      ; housekeeping
[00002173] 8bec             mov ebp,esp   ; housekeeping
[00002175] 6872210000       push 00002172 ; push DDD
[0000217a] e853f4ffff       call 000015d2 ; call HHH(DDD)
[0000217f] 83c404           add esp,+04
[00002182] 5d               pop ebp
[00002183] c3               ret
Size in bytes:(0018) [00002183]
That you already know that it does prove that DDD correctly
emulated by HHH would never stop running unless aborted
or out-of-memory error
*proves that you are trying to get away with a bald-faced lie*
I really hope that you repent before it is too late.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer