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On 6/19/2024 3:57 AM, joes wrote:Am Tue, 18 Jun 2024 21:30:43 -0500 schrieb olcott:On 6/18/2024 9:16 PM, Richard Damon wrote:On 6/18/24 1:25 PM, olcott wrote:On 6/18/2024 12:06 PM, joes wrote:
Can you accept this?You are confusing the machines with their simulators. No longerAnd then are no longer UTMs, and YES, if a machine based on such amWhen such a UTM has been adapted to only simulate the first ten states
modifed UTM (so it is no long a UTM) when the UTM stops simulating,
we can not say the input halted, nor can we say it didn't halt.
of its input TMD, then every simulated TMD with more than ten states
did not terminate normally.
simulating has nothing to do with the simulatee. It does not "know" it
is being simulated. That is entirely in the power of the simulator.
Only it can freely choose to simulate more steps. The simulated machine
then proceeds.
The TM does not halt/terminate. The simulator does. Otherwise it wouldI am establishing the notion of abnormal termination for Turing machinesThe not-a-UTM just came to a no-answer state.I have to go one-step-at-a-time with everyone or they get overwhelmed
and leap to the conclusion that I am wrong.
within the standard terms of the art.
No, that part is correct.void DDD()Oh, they can. It's just that repeating states don't halt in a finiteThe answer will be provided by useing an ACTUAL UTM that keeps onYou are stuck on the idea that repeating states cannot be recognized
going, or the direct execution of the machine,
in a finite number of steps.
number of steps.
{
H0(DDD);
}
Richard cannot understand that H0 can determine in a finite
number of steps that DDD correctly emulated by H0 cannot halt in a
finite number of steps. I used to think that he was lying about this. It
seems that the actual case is that I overestimated his skill level.
Now that he says that when Bonita said "Everything correct"The continuation "no further questions allowed" makes it sarcastic. They
he is taking this to mean something is wrong I am back to thinking that
he might be a liar.
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