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On 7/16/2024 3:20 AM, Mikko wrote:Nope, because the finite string doesn't represent a program, and when you include the information that HHH adds to it, it becomes a Halting Program.On 2024-07-15 20:56:21 +0000, olcott said:HHH computes the mapping from this finite string: 558bec6863210000e853f4ffff83c4045dc3
>On 7/15/2024 3:51 PM, joes wrote:>Am Mon, 15 Jul 2024 08:51:14 -0500 schrieb olcott:>On 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.
>
Not at all. The huge mistake of all these years is that
people stupidly expected that HHH to report on the behavior
of its own executing Turing machine.
No, the error is that HHH report on its own behavour instead of the
behaviour specified by its inputs. Nobody expects that your programs
do anything interesting or useful.
>
to non-halting behavior because this finite string
calls HHH(DDD) in recursive simulation.
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