Sujet : Re: Halting Problem: What Constitutes Pathological Input
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theoryDate : 07. May 2025, 09:17:36
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <1a9a7405aee995d4ca56351b940e42be7d66e0ae@i2pn2.org>
References : 1 2 3 4 5 6 7 8
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Tue, 06 May 2025 21:40:14 -0500 schrieb olcott:
On 5/6/2025 6:00 PM, Richard Damon wrote:
On 5/6/25 1:54 PM, olcott wrote:
On 5/6/2025 6:06 AM, Richard Damon wrote:
On 5/5/25 10:29 PM, olcott wrote:
On 5/5/2025 8:06 PM, Richard Damon wrote:
On 5/5/25 11:51 AM, olcott wrote:
When HHH computes the mapping from *its input* to the behavior of
DD emulated by HHH this includes HHH emulating itself emulating
DD. This matches the infinite recursion behavior pattern.
>
And *ITS INPUT*, for the HHH that answers 0, is the representation
of a program
>
Not at all. This has always been stupidly wrong.
The input is actually a 100% perfectly precise sequence of steps.
With pathological self-reference some of these steps are inside the
termination analyzer.
>
Can't be, as the input needs to be about a program, which must, by
the definition of a program, include all its algorithm.
Yes, there are steps that also occur in the termination analyzer, but
they have been effectively copied into the program the input
describes.
What you forget is that the input program INCLUDES as its definiton,
all of the code it uses, and thus the call to the decider it is built
on includes that code into the decider, and that is a FIXED and
DETERMINDED version of the decider, the one that THIS version of the
input is designed to make wrong.
This doesn't change when you hypothosize a different decider looking
at THIS input.
>
*would never stop running unless aborted*
Refers to a hypothetical HHH/DD pair of the same HHH that DD calls
except that this hypothetical HHH never aborts.
>
Right, but a correct simulation of D does halt,
A correct simulation is one that produces the same behaviour as the
direct execution. HHH does not.
How the Hell is breaking the rules specified by the x86 language
possibly correct?
The rule that you may not abort? The rule that you may not simulate
hypothetical code?
-- Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:It is not guaranteed that n+1 exists for every n.
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