Sujet : Re: Halting Problem: What Constitutes Pathological Input
De : dbush.mobile (at) *nospam* gmail.com (dbush)
Groupes : comp.theoryDate : 05. May 2025, 16:58:50
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vvan7q$o4v0$1@dont-email.me>
References : 1 2
User-Agent : Mozilla Thunderbird
On 5/5/2025 11:51 AM, olcott wrote:
On 5/5/2025 10:17 AM, Mr Flibble wrote:
What constitutes halting problem pathological input:
>
Input that would cause infinite recursion when using a decider of the
simulating kind.
>
Such input forms a category error which results in the halting problem
being ill-formed as currently defined.
>
/Flibble
I prefer to look at it as a counter-example that refutes
all of the halting problem proofs.
Which start with the assumption that the following mapping is computable and that (in this case) HHH computes it:
Given any algorithm (i.e. a fixed immutable sequence of instructions) X described as <X> with input Y:
A solution to the halting problem is an algorithm H that computes the following mapping:
(<X>,Y) maps to 1 if and only if X(Y) halts when executed directly
(<X>,Y) maps to 0 if and only if X(Y) does not halt when executed directly
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
https://github.com/plolcott/x86utm
The x86utm operating system includes fully
operational HHH and DD.
https://github.com/plolcott/x86utm/blob/master/Halt7.c
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.
Thus the Halting Problem's "impossible" input
is correctly determined to be non-halting.
Which is a contradiction. Therefore the assumption that the above mapping is computable is proven false, as Linz and others have proved and as you have *explicitly* agreed is correct.