Sujet : Re: DDD correctly emulated by HHH is correctly rejected as non-halting. --- You are not paying attention
De : polcott333 (at) *nospam* gmail.com (olcott)
Groupes : comp.theoryDate : 16. Jul 2024, 15:46:40
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v7614g$19j7l$11@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 7/16/2024 2:18 AM, Mikko wrote:
On 2024-07-15 13:32:27 +0000, olcott said:
On 7/15/2024 2:57 AM, Mikko wrote:
On 2024-07-14 14:48:05 +0000, olcott said:
>
On 7/14/2024 3:49 AM, Mikko wrote:
On 2024-07-13 12:18:27 +0000, olcott said:
>
When the source of your disagreement is your own ignorance
then your disagreement has no actual basis.
>
*You can comprehend this is a truism or fail to*
*comprehend it disagreement is necessarily incorrect*
Any input that must be aborted to prevent the non
termination of HHH necessarily specifies non-halting
behavior or it would never need to be aborted.
>
Disagreeing with the above is analogous to disagreeing
with arithmetic.
>
A lame analogy. A better one is: 2 + 3 = 5 is a proven theorem just
like the uncomputability of halting is.
>
The uncomputability of halting is only proven when the problem
is framed this way: HHH is required to report on the behavior
of an input that was defined to do exactly the opposite of
whatever DDD reports.
No, it is proven about the halting problem as that problem is.
Which is simply a logical impossibility thus no actual
limit to computation more that this logical impossibility:
What time is it (yes or no)?
*This is isomorphic the HP decider/input pair*
Can Carol correctly answer “no” to this (yes/no) question? (Hehner:2018:2)
Giving credit where credit is due Richard corrected
a loophole in the original question.
The program that predicts what HHH would say and does the opposite
is just one eample of a program.
It is just like a Liar Paradox input to a True(L, x) predicate.
The correct answer is INVALID INPUT.
When HHH is defined such that an input that was defined to
do the opposite of whatever HHH reports can never reach this
point in its execution trace then the prior halting problem
proof has been defeated.
From a programmer's point of view, if we apply an
interpreter to a program text that includes a call
to that same interpreter with that same text as
argument, then we have an infinite loop. A halting
program has some of the same character as an interpreter:
it applies to texts through abstract interpretation.
Unsurprisingly, if we apply a halting program to a
program text that includes a call to that same halting
program with that same text as argument, then we have
an infinite loop. (Hehner:2011:15)
[5] E C R Hehner. Problems with the Halting Problem, COMPUTING2011 Symposium on 75 years of Turing Machine and Lambda-Calculus, Karlsruhe Germany, invited, 2011 October 20-21; Advances in Computer Science and Engineering v.10 n.1 p.31-60, 2013
https://www.cs.toronto.edu/~hehner/PHP.pdfNo, not anymore that 2 + 3 = 5 is defeated by a 2 that is defined to
shrink to 1 if 3 is added to it.
*Simulating Termination Analyzer H is Not Fooled by Pathological Input D*
https://www.researchgate.net/publication/369971402_Simulating_Termination_Analyzer_H_is_Not_Fooled_by_Pathological_Input_D --
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer