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On 4/2/2025 1:47 PM, Fred. Zwarts wrote:Really? Why doesn't the direct execution of that input, when we include the HHH that you have defined?Op 02.apr.2025 om 17:55 schreef olcott:On 4/2/2025 9:14 AM, joes wrote:Am Mon, 31 Mar 2025 16:26:58 -0500 schrieb olcott:>On 3/31/2025 2:10 PM, Fred. Zwarts wrote:>Op 31.mrt.2025 om 20:16 schreef olcott:Why are you not passing DDD as input? Why do you not call what you're>A simulating termination analyzer is always correct to abort theBut the input is halting, as proven by direct execution.
simulation and reject the input as non-halting when-so-ever this input
would otherwise prevent itself from halting.
>
Something other than the input is halting.
HHH1(DDD) shows the same behavior as the direct execution.
HHH(DDD) shows the behavior of the actual input.
doing HHH(HHH(DDD))? What is the difference in what is passed to HHH1?
>
This seems to be above your level of technical competence.
>
_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]
>
Anyone understanding the above code where HHH
emulates DDD according to the semantics of the
x86 language knows that this DDD (not some
other different DDD) cannot possibly reach its
own final halt state.Yes it fails to reach the end of the simulation of a program that according to the x86 semantics has an end as proven by direct execution.In other words you don't hardly know the x86
language at all.
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