On 2025-11-08, olcott <polcott333@gmail.com> wrote:
On 11/8/2025 1:58 PM, Kaz Kylheku wrote:
On 2025-11-08, olcott <polcott333@gmail.com> wrote:
On 11/8/2025 12:10 PM, Kaz Kylheku wrote:
On 2025-11-08, olcott <polcott333@gmail.com> wrote:
On 11/8/2025 5:49 AM, Richard Heathfield wrote:
On 08/11/2025 09:27, Tristan Wibberley wrote:
On 08/11/2025 04:57, Richard Heathfield wrote:Olcott's situation has no solution.
On 07/11/2025 20:57, olcott wrote:
<snip>
In my specific case D simulated by H specifies a different >>>>>>>>>> sequence of steps than D executed from main because they
are executed in different contexts.
If you give the 'decider' licence to choose an execution context, you >>>>>>>>> can write a universal decider easily:
int H(int (*d)())
{
-a-a return 1; /* in H's context, all programs halt */
}
But the decider is *not* granted that licence.
Olcott's situation doesn't require that licence in order to be solved. >>>>>>>
D simulated by H cannot possibly reach its own
simulated "return" statement final halt state
thus the input to H(D) specifies a non-halting
sequence of configurations.
This is true of the above H, which returns 1 (accept).
It is not true of any H that returns 0 for D,
no matter how that 0 is calculated.
You only think this because you wrongly reject the idea that the
simulation is not finished when it is aborted by H.
/Neglecting to simulate/ D's termination is not the same thing
as D not having one.
D simulated by H cannot possibly have an
"D simulated by H" is literally not a thing. D is simulated by
a simulator, which doesn't care whether it is driven by
events from H, or elsewhere.
All correct simulations of D show halting.
Only when you dishonestly ignore that we are only
examining the case where D calls its own simulator.
D never calls its own simulator; there is no such thing.
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