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On 4/26/2025 3:56 AM, Fred. Zwarts wrote:This behaviour of HHH blocks the analysis of the behaviour of DD.Op 25.apr.2025 om 20:42 schreef olcott:_DD()On 4/25/2025 8:36 AM, Richard Damon wrote:Indeed, HHH is programmed such that it fails to reach the endOn 4/25/25 9:08 AM, olcott wrote:>On 4/24/2025 7:07 PM, Richard Damon wrote:>On 4/24/25 7:58 PM, olcott wrote:>On 4/24/2025 6:14 PM, Richard Damon wrote:>On 4/24/25 5:13 PM, olcott wrote:>On 4/24/2025 5:59 AM, Richard Damon wrote:>On 4/23/25 11:22 PM, polcott333 wrote:>On 4/23/2025 9:41 PM, Richard Damon wrote:>On 4/23/25 11:32 AM, olcott wrote:>On 4/23/2025 6:25 AM, joes wrote:>Am Tue, 22 Apr 2025 13:51:48 -0500 schrieb olcott:>On 4/22/2025 1:07 PM, Fred. Zwarts wrote:>Op 22.apr.2025 om 18:28 schreef olcott:On 4/22/2025 7:57 AM, joes wrote:Am Tue, 15 Apr 2025 15:44:06 -0500 schrieb olcott:No, DD halts (when executed directly). HHH is not a halt decider, not even>>libx86emu <is> a correct x86 processor and does emulate its inputsYou continue to stupidly insist that int sum(int x, int y) {return xWhat else is it missing that the processor uses to execute it?
+ y; }
returns 7 for sum(3,2) because you incorrectly understand how these
things fundamentally work.
>
It is stupidly wrong to expect HHH(DD) report on the direct
execution of DD when you are not telling it one damn thing about
this direct execution.
>
correctly.
The key thing here is that Olcott consistently does not understand that
HHH is given a finite string input that according to the semantics of
the x86 language specifies a halting program,
That is stupidly incorrect.
for DD only.
>People here stupidly assume that the outputs are not required toBut the direct execution of DD is computable from its description.
correspond to the inputs.
>
Not as an input to HHH.
But neither the "direct execution" or the "simulation by HHH" are "inputs" to HHH. What is the input is the representation of the program to be decided on.
>When HHH computes halting for DD is is only allowed>
to apply the finite string transformations specified
by the x86 language to the machine code of DD.
It is only ABLE to apply them.
>
The input to HHH(DD) does specify the recursive emulation
of DD including HHH emulating itself emulating DD when
one applies the finite string transformation rules of the
x86 language to THE INPUT to HHH(DD).
Yes, the input specifies FINITE recusive PARTIAL emulation, as the HHH that DD calls will emulate only a few instructions of DD and then return,
>
*You are technically incompetent on this point*
When the finite string transformation rules of the
x86 language are applied to the input to HHH(DD)
THIS DD CANNOT POSSIBLY REACH ITS FINAL HALT STATE
not even after an infinite number of emulated steps.
Sure it does, just after the point that HHH gives up on those transformation and aborts its (now incorrect) emulation of the input.
>
THAT IS COUNTER FACTUAL !!!
>
The directly executed DD has zero recursive invocations.
DD emulated by HHH has one recursive invocation.
>
Did you know that zero does not equal one?
>
But the direct execution DOES have a recursiove invocation, as DD calls HHH(DD) that emulated DD, just like the directly exeucted HHH will emulate DD calling HHH(DD).
>
The call from the directly executed DD to HHH(DD)
immediately returns and DD reaches its final halt state.
No it doesn't,
I will say that more accurately.
The call from the directly executed DD returns.
>
The call from DD emulated by HHH to HHH(DD)
(according to the finite string transformation
rules of the x86 language) CANNOT POSSIBLY RETURN.
[00002133] 55 push ebp ; housekeeping
[00002134] 8bec mov ebp,esp ; housekeeping
[00002136] 51 push ecx ; make space for local
[00002137] 6833210000 push 00002133 ; push DD
[0000213c] e882f4ffff call 000015c3 ; call HHH(DD)
[00002141] 83c404 add esp,+04
[00002144] 8945fc mov [ebp-04],eax
[00002147] 837dfc00 cmp dword [ebp-04],+00
[0000214b] 7402 jz 0000214f
[0000214d] ebfe jmp 0000214d
[0000214f] 8b45fc mov eax,[ebp-04]
[00002152] 8be5 mov esp,ebp
[00002154] 5d pop ebp
[00002155] c3 ret
Size in bytes:(0035) [00002155]
When any HHH emulates DD according to the finite
string transformation rules specified by the x86
language (the line of demarcation between correct
and incorrect emulation) no emulated DD can possibly
reach its final halt state and halt.
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