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On 6/28/2024 8:14 AM, joes wrote:Am Thu, 27 Jun 2024 12:30:38 -0500 schrieb olcott:
Where do you disagree?When this is construed as non-halting criteria then simulatingTo the caller DDD, which then returns to its own caller H0, which
termination analyzer H0 is correct to reject these inputs as
non-halting by returning 0 to its caller.
returns „halting” to main… hold on.
Simulating termination analyzers must report on the behavior thatH0 must not report on itself, only on DDD. Which you’ve proven halts.
their finite string input specifies thus H0 must report that DDD
correctly emulated by H0 remains stuck in recursive simulation.
We don’t care how H0 deviates (i.e. is incorrect) in its simulation.
That would be main {H0(H0(DDD))}.
The behavior of the directly executed DDD() is irrelevant because thatWhat is the difference here?
is not the behavior of the input.
Deciders compute the mapping fromAnd should get the right answer.
their actual finite string input to an output by a sequence of finite
string transformations.
In this case the sequence is the line-by-line execution trace of theNo, the sequence is the behaviour of DDD, period.
behavior of DDD correctly emulated by HHH.
The behavior of this input must include and cannot ignore the recursiveYes, and the behaviour of H0 is that it produces the exact same behaviour
emulation specified by the fact that DDD is calling its own emulator.
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