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On 8/9/2024 4:03 AM, Mikko wrote:No, since all the emulator that only emulated a finite number of steps did not complete the job, and thus did NOT determine the actual mapping of that input, which is determined by RUNNING that DDD, or COMPLETELY emulating it, which will see the DDD call HHH(DDD) which will after emulationg that finite number of instructions return to DDD which will halt.On 2024-08-08 13:18:34 +0000, olcott said:Here is something that literally does not mean anything:
>void DDD()>
{
HHH(DDD);
return;
}
>
Each HHH of every HHH that can possibly exist definitely
*emulates zero to infinity instructions correctly* In
none of these cases does the emulated DDD ever reach
its "return" instruction halt state.
The ranges of "each HHH" and "every HHH" are not defined above
so that does not really mean anything.
>
"0i34ine ir m0945r (*&ubYU I*(ubn)I*054 gfdpodf["
"Colorless green ideas sleep furiously"
https://en.wikipedia.org/wiki/Colorless_green_ideas_sleep_furiously
Has lots of meaning, that does not totally fit together coherently.
I defined an infinite set of HHH x86 emulators.
I stipulated that each member of this set emulates
zero to infinity instructions of DDD.
*I can't say it this way without losing 90% of my audience*
Each element of this set is mapped to one element of the
set of non-negative integers indicating the number of
x86 instructions of DDD that it emulates.
*This one seems to be good*Right, but any number emulated that is less than the number of instructions that particular input needs to have run to reach a final state is just INCORRECT, since all the HHH that abort DO create an input that will halt in a finite number of steps, just at a value of N bigger than the number of steps that HHH emulated.
Each element of this set corresponds to one element of
the set of positive integers indicating the number of
x86 instructions of DDD that it emulates.
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