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Op 03.sep.2024 om 00:22 schreef olcott:It does this correctly yet beyond your intellectual capacity.On 9/2/2024 12:52 PM, Fred. Zwarts wrote:It is not DDD. It is HHH that has the problem when trying to simulate itself.Op 02.sep.2024 om 18:38 schreef olcott:>A halt decider is a Turing machine that computes>
the mapping from its finite string input to the
behavior that this finite string specifies.
>
If the finite string machine string machine
description specifies that it cannot possibly
reach its own final halt state then this machine
description specifies non-halting behavior.
>
A halt decider never ever computes the mapping
for the computation that itself is contained within.
>
Unless there is a pathological relationship between
the halt decider H and its input D the direct execution
of this input D will always have identical behavior to
D correctly simulated by simulating halt decider H.
>
*Simulating Termination Analyzer H Not Fooled by Pathological Input D*
https://www.researchgate.net/ publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D
>
A correct emulation of DDD by HHH only requires that HHH
emulate the instructions of DDD** including when DDD calls
HHH in recursive emulation such that HHH emulates itself
emulating DDD.
Indeed, it should simulate *itself* and not a hypothetical other HHH with different behaviour.
If HHH includes code to see a 'special condition' and aborts and halts, then it should also simulate the HHH that includes this same code and
>
DDD has itself and the emulated HHH stuck in recursive emulation.
>
void DDD()
{
HHH(DDD);
return;
}
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