Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--

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Sujet : Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--
De : polcott2 (at) *nospam* gmail.com (olcott)
Groupes : comp.theory sci.logic
Date : 08. Mar 2024, 05:39:51
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <use1a7$1gd96$1@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
User-Agent : Mozilla Thunderbird
On 3/7/2024 8:45 PM, Richard Damon wrote:
On 3/7/24 6:05 PM, olcott wrote:
On 3/7/2024 7:36 PM, immibis wrote:
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);
}
>
 Nope, only if H actually does (or attempts to do) a complete simulation.
 
My point is that the infinite loop has no effect what-so-ever on
the ability of a termination analyzer to correctly analyze an
input such as either version of D.

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
can possibly do correctly determine that they themselves are called
in recursive simulation.
This same extra input that H(D,D) and H1(D,D) has is what enables
them to correctly compute different values.

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.
 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.
I presume nothing and proved that Olcott machines have a power
that Turing machines do not have. They can (just like H and H1)
correctly determine when they themselves are called in recursive
simulation.
*It has taken me about two years to put this intuition into words*
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer

Date Sujet#  Auteur
7 Mar 24 * Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--10immibis
7 Mar 24 +* Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--6olcott
8 Mar 24 i`* Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--5immibis
8 Mar 24 i `* Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--4olcott
8 Mar 24 i  `* Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--3Richard Damon
8 Mar 24 i   `* Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--2olcott
8 Mar 24 i    `- Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--1Richard Damon
7 Mar 24 `* Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--3Mike Terry
7 Mar 24  `* Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--2olcott
7 Mar 24   `- Re: Refutation of the Peter Linz Halting Problem proof 2024-03-05 --partial agreement--1Richard Damon

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