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Am Mon, 24 Mar 2025 17:43:14 -0500 schrieb olcott:They never do, they only report on the behavior thatOn 3/24/2025 4:49 PM, André G. Isaak wrote:On 2025-03-24 14:11, olcott wrote:On 3/24/2025 12:35 PM, dbush wrote:On 3/24/2025 12:44 PM, olcott wrote:On 3/24/2025 10:14 AM, dbush wrote:On 3/24/2025 11:03 AM, olcott wrote:On 3/24/2025 6:23 AM, Richard Damon wrote:On 3/23/25 11:09 PM, olcott wrote:UTMs do.The whole point of this post is to prove that no Turing machine everThe HHH you implemented is computing *a* computable function, butgiven an input of the function domain it can return theWhich includes the machine code of DDD, the machine code of HHH,DDD is a semantically and syntactically correct finite stirng ofIt is impossible for HHH compute the function from the directWHy isn't DDD made into the correct finite string?i
execution of DDD because DDD is not the finite string input
basis from which all computations must begin.
https://en.wikipedia.org/wiki/Computable_function
the x86 machine language.
and the machine code of everything it calls down to the OS level.
>
>Which is another way of saying that HHH can't determine that DDDThat seems to be your own fault.DDD emulated by HHH directly causes recursive emulation because it
The problem has always been that you want to use the wrong string
for DDD by excluding the code for HHH from it.
calls HHH(DDD) to emulate itself again. HHH complies until HHH
determines that this cycle cannot possibly reach the final halt
state of DDD.
halts when executed directly.
corresponding output.
Computable functions are only allowed to compute the mapping from
their input finite strings to an output.
it's not computing the halting function:
reports on the behavior of the direct execution of another Turing
machine.
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