Op 26.jul.2024 om 15:22 schreef olcott:
On 7/26/2024 1:53 AM, Fred. Zwarts wrote:
Op 26.jul.2024 om 03:49 schreef olcott:
If you understand the x86 language and can't tell how DDD
emulated by HHH differs from DDD emulated by HHH1 by the
following then you are probably lying about understanding
the x86 language.
We understand it perfectly. HHH cannot possibly simulate itself
correctly.
You are too stupid to know that a non-halting computation
cannot be emulated to completion because completion does
not exist.
The non-halting behaviour is only in your dreams. It is irrelevant,
because HHH halts when it aborts. Remember, HHH is simulating *itself*,
a halting program, not another non-halting simulator that does not abort
and does not halt.
On this stupid basis you say that because HHH does not simulate
DDD to a completion that does not exist that DDD is simulated
incorrectly. Unlike pure ad hominem rebuttals I point out how
and why what you say is stupidly incorrect.
Since you do not understand that there is a difference between two and infinite, you think that there is an infinite recursion (non-halting)
that must be halted. But the non-halting behaviour is just your dream.
You are unable to see the difference between your dream and reality.
The semantics of the x86 language of this code show, when directly
executed, that there is no non-halting behaviour. HHH halts after two
cycles. This semantics does not change when we give the exact same code
to a correct simulator, like HHH1, which also shows that it halts after
two cycles.
The semantics of the x86 language does not change when we give the exact
same code to an incorrect simulator, like HHH, but HHH fails to properly simulate this code up to the end, because it aborts the simulation after
one cycle, when the simulation still has one cycle to go, proving that
the simulation is incomplete and incorrect. It shows only the first part
of the behaviour of HHH, not its full behaviour.
Two is different from infinite. One is different from infinite.
If you think this is stupid, it tells us more about you, than about this reasoning. You do not have the mental intelligence to even consider the possibility that you are wrong, therefore, if you cannot point to an
error, you just ignore it using the excuse that it is stupid or untrue
what people tell you.
HHH is simply unable to decide about finite recursions.
void Finite_Recursion (int N) {
if (N > 0) Finite_Recursion (N - 1);
}
This is a similar finite recursion as that of HHH, which halts after two cycles.
It decides after two recursions that there is an infinite recursion,
which is incorrect.
HH1 does a correct simulation, but HHH does an incomplete and
therefore incorrect simulation. The incomplete simulation does not
show the full behaviour of DDD.
Since skipping x86 code is against the semantics of the x86 language,
it is clear where the error is.
*I did annotate it a little better this time*
typedef void (*ptr)();
int HHH(ptr P);
int HHH1(ptr P);
void DDD()
{
HHH(DDD);
}
int main()
{
HHH1(DDD);
}
*You really don't need to know one damn thing else besides this*
*You really don't need to know one damn thing else besides this*
*You really don't need to know one damn thing else besides this*
All that you have to know is that HHH and HHH1 are x86 emulators
and that HHH sees that same repeated state (first four lines of DDD)
that anyone knowing the x86 language can see.
_DDD()
[00002177] 55 push ebp
[00002178] 8bec mov ebp,esp
[0000217a] 6877210000 push 00002177 ; push DDD
[0000217f] e853f4ffff call 000015d7 ; call HHH
[00002184] 83c404 add esp,+04
[00002187] 5d pop ebp
[00002188] c3 ret
Size in bytes:(0018) [00002188]
_main()
[00002197] 55 push ebp
[00002198] 8bec mov ebp,esp
[0000219a] 6877210000 push 00002177 ' push DDD
[0000219f] e863f3ffff call 00001507 ; call HH1
[000021a4] 83c404 add esp,+04
[000021a7] 33c0 xor eax,eax
[000021a9] 5d pop ebp
[000021aa] c3 ret
Size in bytes:(0020) [000021aa]
machine stack stack machine assembly
address address data code language
======== ======== ======== ========= =============
[00002197][001037fb][00000000] 55 push ebp
[00002198][001037fb][00000000] 8bec mov ebp,esp
[0000219a][001037f7][00002177] 6877210000 push 00002177 ; push DDD
[0000219f][001037f3][000021a4] e863f3ffff call 00001507 ; call HHH1
New slave_stack at:10389f
// emulates 1st instance of DDD that calls HHH(DDD)
Begin Local Halt Decider Simulation Execution Trace Stored at:1138a7 >>> [00002177][00113897][0011389b] 55 push ebp
[00002178][00113897][0011389b] 8bec mov ebp,esp
[0000217a][00113893][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][0011388f][00002184] e853f4ffff call 000015d7 ; call HHH
New slave_stack at:14e2c7
// emulates 2nd instance of DDD that calls HHH(DDD)
Begin Local Halt Decider Simulation Execution Trace Stored at:15e2cf >>> [00002177][0015e2bf][0015e2c3] 55 push ebp
[00002178][0015e2bf][0015e2c3] 8bec mov ebp,esp
[0000217a][0015e2bb][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][0015e2b7][00002184] e853f4ffff call 000015d7 ; call HHH
New slave_stack at:198cef
// emulates 3rd instance of DDD that calls HHH(DDD)
[00002177][001a8ce7][001a8ceb] 55 push ebp
[00002178][001a8ce7][001a8ceb] 8bec mov ebp,esp
[0000217a][001a8ce3][00002177] 6877210000 push 00002177 ; push DDD
[0000217f][001a8cdf][00002184] e853f4ffff call 000015d7 ; call HHH
Local Halt Decider: Infinite Recursion Detected Simulation Stopped
This is the error in HHH. HHH is programmed to print after two cycles
that there is an infinite recursion, which is not true, after which it
aborts and halts, making the simulation incorrect.
HHH is simply unable to decide about finite recursions.
void Finite_Recursion (int N) {
if (N > 0) Finite_Recursion (N - 1);
}
It decides after N recursions that there is an infinite recursion,
which is incorrect. It does not see the difference between an finite
and an infinite recursion.
// returns to 1st instance of DDD emulated by HHH1
[00002184][00113897][0011389b] 83c404 add esp,+04
[00002187][0011389b][000015bc] 5d pop ebp
[00002188][0011389f][0003a980] c3 ret
// return to main
[000021a4][001037fb][00000000] 83c404 add esp,+04
[000021a7][001037fb][00000000] 33c0 xor eax,eax
[000021a9][001037ff][00000018] 5d pop ebp
[000021aa][00103803][00000000] c3 ret
Number of Instructions Executed(352831) == 5266 Pages
Olcott's psychology is intriguing. At the one hand he is crying for
help. He has some ideas which he cannot prove. Therefore he is begging
the exports to help him with a proof.
When the experts prove that there are errors in his idea and show him
a way to improve his idea, he does not have the mental intelligence to
even consider the possibility that there is something incorrect in his
ideas. He cannot show any error in their proofs, but he does not want
to accept it.
Therefore he uses the excuse that the experts must be lying in order
to be able to ignore the proofs that his ideas are wrong.
Then he keeps repeating his proven incorrect ideas, without any
evidence, probably hoping that if it is repeated often enough, it will
become true.
That is not how logic works.
No matter how much olcott wants it to be correct, or how many times
olcott repeats that it is correct, it does not change the fact that
these ideas are incorrect.
--- SoupGate-Win32 v1.05
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