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On 3/7/24 6:05 PM, olcott wrote:My point is that the infinite loop has no effect what-so-ever onOn 3/7/2024 7:36 PM, immibis wrote:Nope, only if H actually does (or attempts to do) a complete simulation.On 7/03/24 18:14, olcott wrote:>It is an easily fact that a correct and complete simulation of>
the input to H(D,D) and the input to Ĥ.H ⟨Ĥ⟩ ⟨Ĥ⟩ cannot possibly halt.
It can if H(D,D) returns 0.
D simulated by H cannot possibly halt even if we simplify it to this:
>
void D(void (*x)())
{
H(x, x);
}
>
If H actually aborts its simulation and return 0, then a CORRECT simulation of the input to H will halt.It is still the same paradox for a different reason.
So, you are stuck, because Computations are based on deterministic algorithms. Either H's algorithm WILL continue to simulate through every thing it sees in its simulation of D, and then neither machine will halt, so H doesn't answer and is wrong.Not at all. Olcott machines can do something that no Turing machine
Or H gives up and decides it needs to stop to give an answer, but then, not seeing what D will do, is forced to "Guess", (and will always guess this input the same way) and when we actually run D, it gets to see how H guesses, and makes it wrong.I presume nothing and proved that Olcott machines have a power
You keep on presuming that the H in D will be somehow different then the H that is deciding on it, which is just a LIE.
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