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On 10/6/2024 12:29 PM, Richard Damon wrote:No, that isn't ignoring it, but taking into account that since HHH is defined to be a specific program, it has specific behavior.On 10/6/24 1:07 PM, olcott wrote:gets to ignore the fact that DDD was defined toOn 10/6/2024 11:59 AM, Richard Damon wrote:>On 10/6/24 8:39 AM, olcott wrote:>>>
DDD emulated by each corresponding HHH that can possibly
exist never returns. Each of these HHH emulators that does
return 0 correctly reports the above non-halting behavior.
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No, the DDD return (if the HHH(DDD) gives an answer), just after the HHH that emulated them gave up.
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DDD emulated by each corresponding HHH that can possibly
exist never returns.
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DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
DDD emulated by each corresponding HHH
Which, as you have been told but seems to be above your head means that the execution of DDD,
have a pathological relationship with HHH that
HHH cannot ignore.
Everyone that has been disagreeing with me insistsNope, it must take the behavior of the actual HHH into account.
that a correct emulation of DDD by HHH must make
sure to ignore what the DDD/HHH code specifies.
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