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On 5/2/24 2:46 PM, olcott wrote:The infinite set of every possible implementation of H includesOn 5/2/2024 6:03 AM, Richard Damon wrote:Then simulate EXACTLY 1 step and abort.On 5/2/24 12:21 AM, olcott wrote:>On 5/1/2024 7:28 PM, Richard Damon wrote:>On 5/1/24 11:51 AM, olcott wrote:>Every D simulated by H that cannot possibly stop running unless>
aborted by H does specify non-terminating behavior to H. When
H aborts this simulation that does not count as D halting.
Which is just meaningless gobbledygook by your definitions.
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It means that
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int H(ptr m, ptr d) {
return 0;
}
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Your H is not simulating D at all thus not
"Every D simulated by H" quoted above
Yes it is, it is just aborting the simulation before it started.
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D is never simulated by H is not the same as D is simulated by H.
D is simulated by H entails that 1 to N steps are simulated and
then the simulation is aborted or the simulation is never aborted.
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It makes all but the most trivial programs "non-halting."--
Again, a TOY.,
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