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Am Mon, 15 Jul 2024 08:51:14 -0500 schrieb olcott:Not at all. The huge mistake of all these years is thatOn 7/15/2024 3:37 AM, Mikko wrote:Same difference.On 2024-07-15 03:41:24 +0000, olcott said:No that is wrong. The finite string must encode a Turing machine.On 7/14/2024 9:04 PM, Richard Damon wrote:>On 7/14/24 9:27 PM, olcott wrote:Turing machines only operate on finite strings they do not operate on>>
Any input that must be aborted to prevent the non termination of
simulating termination analyzer HHH necessarily specifies
non-halting behavior or it would never need to be aborted.
Excpet, as I have shown, it doesn't.
Your problem is you keep on ILEGALLY changing the input in your
argument because you have misdefined what the input is.
The input to HHH is ALL of the memory that it would be accessed in a
correct simulation of DDD, which includes all the codd of HHH, and
thus, if you change HHH you get a different input.
If you want to try to claim the input is just the bytes of the
function DDD proper then you are just admitting that you are nothing
more than a lying idiot that doesn't understand the problem,
other Turing machines *dumbo*
That's right. But the finite string can be a description of a Turing
machine.
It is a despicable lie that it even be called "undecidable".Which is that other TM.That way a Turing machine can say someting about another TuringNot exactly. It can only report on the behavior that the input finite
machine,
string specifies.
Not true. Some interesting questions are undecidable.even simulate its complete execution. Or it can count something simpleAll of the questions that a TM cannot answer are logical impossibilities
like the number of states or the set of symbols that the described
Turing machine may write but not erase. But there are questions that no
Turing machine can answer from a description of another Turing machine.
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