From Newsgroup: comp.lang.c++
On 8/21/2025 10:44 AM, Kaz Kylheku wrote:
On 2025-08-21, olcott <polcott333@gmail.com> wrote:
On 8/20/2025 10:33 PM, Richard Heathfield wrote:
The conventional proof does not require the existence of the input you
describe,
Cite your sources.
I have been studying this for 22 years
and never saw a proof that did not require
an input to do or say the opposite of what
its decider says.
You've been studying it wrong. The input contains its own copy of a
certain decider. Which decider it contains does not vary with the
decider being applied to that input.
The embedded decider may be a clean-room implementation of the
algorithm description developed by the author of the input,
based on a description of the decider.
(Needless to say, it is not valid for a decider algorithm to have
elements like "look at your own code bytes to share mutable
state with another running implementation of the decider".)
I know that. You taught me that.
I don't currently know how to change my code so that
HHH can see the recursive simulation execution traces
of DD without the use of static data.
*From the bottom of page 319 has been adapted to this*
https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf
Turing Machine Linz -n applied to its own machine description rf?-nrf-
-n.q0 rf?-nrf- reo* -n.embedded_H rf?-nrf- rf?-nrf- reo* -n.reR,
-n.q0 rf?-nrf- reo* -n.embedded_H rf?-nrf- rf?-nrf- reo* -n.qn
*When -n.embedded_H is based on a UTM*
*Repeats until aborted*
(a) -n copies its input rf?-nrf-
(b) -n invokes embedded_H rf?-nrf- rf?-nrf-
(c) embedded_H simulates rf?-nrf- rf?-nrf-
If -n.embedded_H cannot possibly see the repeating
state then we know something we never knew before:
that -n.embedded_H rf?-nrf- rf?-nrf- reo* -n.qn would be correct.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer
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