olcott
unread,Nov 19, 2021, 2:26:58 PM11/19/21You do not have permission to delete messages in this group
Either email addresses are anonymous for this group or you need the view member email addresses permission to view the original message
to
On 11/19/2021 12:31 PM, André G. Isaak wrote:
> On 2021-11-19 11:06, olcott wrote:
>> On 11/19/2021 11:46 AM, André G. Isaak wrote:
>>> On 2021-11-19 10:32, olcott wrote:
>
>>>> The input to (H,P) never halts P(P) halts.
>>>> Here are the divergent execution sequences at the C level:
>>>>
>>>> int main(){ H(P,P); }
>>>> (1) main()
>>>> (2) calls H(P,P) that simulates the input to H(P,P)
>>>> (3) that calls H(P,P) which aborts its simulation of P(P) and
>>>> returns to
>>>> (4) main().
>>>>
>>>> int main(){ P(P); }
>>>> (a) main() calls P(P) that
>>>> (b) calls H(P,P) that simulates the input to H(P,P)
>>>> (c) that calls H(P,P) which aborts its simulation of P(P) and
>>>> returns to
>>>> (d) P(P) that returns to main()
>>>>
>>>> (1) diverges from (a) causing (4) to diverge from (d)
>>>
>>> And a halt decider is concerned only with the SECOND case, not the
>>> first.
>>>
>>
>> The halt decider is concerned with the execution trace of a sequence
>> of instructions other than the actual execution trace that is
>> specified by actually executing or simulating its actual input?
>
> Your question makes no sense.
It is proven that the execution trace of P(P) and the execution trace of
the input to H(P,P) are not the same and that this difference results in
different behavior.
int main(){ H(P,P); }
(1) main()
(2) calls H(P,P) that simulates the input to H(P,P)
(3) that calls H(P,P) which aborts its simulation of P(P) and returns to
(4) main().
int main(){ P(P); }
(a) main() calls P(P) that
(b) calls H(P,P) that simulates the input to H(P,P)
(c) that calls H(P,P) which aborts its simulation of P(P) and returns to
(d) P(P) that returns to main()
P(P) has a whole other invocation of P(P) prepended to the execution
trace of the input to H(P,P). It is proven that this different execution
trace derives opposite halting behavior between P(P) and the input to
H(P,P).
>
> int main() { P(P); } *is* actually executing the input to int main() {
> H(P,P); }. In int main() { H(P,P); } it is *not* being executed; it is
> being passed as a parameter to another function.
>
> The halting problem asks whether a given computation will halt *when it
> is performed*.
>
> When you execute H(H_Hat, H_Hat) the computation H_Hat(H_Hat) is *not*
> being performed. It is having a question asked about it. That's true
> even if H makes use of partial simulation.
>
> The computation is being performed only when you actually run
> H_Hat(H_Hat), not when you provide a description of it to another Turing
> Machine.
>
> André
>
--
Copyright 2021 Pete Olcott
Talent hits a target no one else can hit;
Genius hits a target no one else can see.
Arthur Schopenhauer