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On 6/30/2024 3:42 AM, joes wrote:Am Sat, 29 Jun 2024 15:03:02 -0500 schrieb olcott:On 6/29/2024 2:44 PM, joes wrote:Am Fri, 28 Jun 2024 14:28:20 -0500 schrieb olcott:On 6/28/2024 2:18 PM, joes wrote:Am Fri, 28 Jun 2024 12:53:46 -0500 schrieb olcott:On 6/28/2024 12:41 PM, joes wrote:Thanks for leaving the unanswered questions in place, though I’d
rather have you answer them.
But it sees that the inner one would abort, so it can let it end normally.The outer HHH meets its abort criteria one execution trace sooner thanIt should abort, just like the outer one.Why doesn’t the first recursive H return?
the next inner one because HHH needs to see two complete execution
traces before its abort criteria has been met.
As soon as the outer HHH sees the inner one complete one full executionSame as the inner does, so the inner one does not in fact repeat.
trace then the outer one has its abort criteria.
Repeat.No, I mean: why does the inner simulator repeat instead of aborting,
the same as the outer one does?
Why are they aborted?My point is: all recursive calls both enter and detect aThe inner ones always see one less execution trace than the next outer
repeating state.
one, thus could only meet their abort criteria after they have already
been aborted.
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