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On Wed, 11 Jun 2025 18:21:37 -0400, Richard Damon wrote:On 6/11/25 2:21 PM, Mr Flibble wrote:On Tue, 10 Jun 2025 23:15:51 -0400, Richard Damon wrote:On 6/10/25 3:05 PM, Mr Flibble wrote:On Tue, 10 Jun 2025 14:53:47 -0400, Richard Damon wrote:On 6/10/25 1:22 PM, olcott wrote:On 6/10/2025 2:33 AM, Mikko wrote:On 2025-06-09 21:14:58 +0000, olcott said:
[AI slop]But it will never "return" because it is infinitely recursive; the>
simulation is aborted and a halting result if non-halting is
returned elsewhere.
So, you have a problem, either you don't have a correct simulation to
show you got the right answer, or you don't answer.
That is the problem with trying to have the decider itself be two
contradictory entities.
A correct simulator can not be a correct decider it the input is
actually non-halting.
There seems to be some mental block about the fact that the
DEFINITION of this sort of decider is that:
H(M) returns 1 if UTM(M) halts, and H(M) returns 0 if UTM(M) will
never halt
If you try to combine the the UTM and H into one program that it can
NEVER correctly return 0, as it can only return 0 if it never halt
(and thus can't return a value)
You are wrong. An SHD does not have to simulate an algorithm to
completion if it determines non-halting early BY ANALYSIS.
I didn't say it needed to. But it needs to determine what such a
simulation will do.
In fact, as I said, if the input IS non-halting, it can't be both the
required simulator and the decider, so it is logically inconsistent to
say that it is the simulation by the decider that defines the result.
t seems you have fallen for Olcott's insanity.
Would you like this analysis as a downloadable text file?I wouldn't like this posted at all, but at least mark it as generated.
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