Re: Halting Problem: What Constitutes Pathological Input

On 5/6/2025 10:43 PM, olcott wrote:

On 5/6/2025 9:30 PM, Richard Damon wrote:

On 5/6/25 9:45 PM, olcott wrote:

On 5/6/2025 5:43 PM, Richard Damon wrote:

On 5/6/25 11:10 AM, olcott wrote:

On 5/6/2025 4:26 AM, Mikko wrote:

On 2025-05-05 18:14:25 +0000, olcott said:
>

On 5/5/2025 11:16 AM, dbush wrote:

On 5/5/2025 12:13 PM, Mr Flibble wrote:

On Mon, 05 May 2025 11:58:50 -0400, dbush wrote:
>

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.

>
The category (type) error manifests in all extant halting problem proofs
including Linz.  It is impossible to prove something which is ill- formed
in the first place.
>
/Flibble

>
All algorithms either halt or do not halt when executed directly. Therefore the problem is not ill formed.

>
When BOTH Boolean RETURN VALUES are the wrong answer
THEN THE PROBLEM IS ILL-FORMED. Self-contradiction must
be screened out as semantically incorrect.

>
Irrelevant. One of the boolean values (the one not returned) is the
right one as can be determined e.g. with an UTM.
>

You only get something that appears that way when a false assumption is made, namely that the halting function is computable.

>
The mapping from the input HHH(DD) finite string of
machine code to DOES SPECIFY RECURSIVE EMULATION
THAT WOULD PREVENT DD FROM EVER HALTING.

>
No, it does not. HHH returns 0 and DD halts.
>

>
You can't show the detailed steps of the execution
trace of DD emulated by HHH (according to the rules
of the x86 language) where DD halts because you are wrong.

>
Because a trace of DD correctly emulatd by HHH doesn't exist as HHH doesn't correctly emulate DD
>

>
Show what the execution trace of DD emulated by HHH
according to the rules of the x86 language should be

>
Since HHH doesn't actually do that, your question is void.
>

 You say it is wrong unless you are lying
you can point out what the correct steps should be.
 

The steps that UTM / HHH1 takes that HHH does not, as your are on record as admitting is correct:
On 5/6/2025 5:17 PM, dbush wrote:
 > On 5/6/2025 5:03 PM, olcott wrote:
 >> On 5/6/2025 3:51 PM, dbush wrote:
 >>> On 5/6/2025 4:46 PM, olcott wrote:
 >>>> On 5/6/2025 3:31 PM, dbush wrote:
 >>>>> Then what is the first instruction emulated by HHH that differs
 >>>>> from the emulation performed by UTM?
 >>>>>
 >>>>
 >>>> HHH1 is exactly the same as HHH except that DD
 >>>> does not call HHH1. This IS the UTM emulator.
 >>>> It does not abort.
 >>>
 >>> Last chance:
 >>>
 >>> What is the first instruction emulated by HHH that differs from the
 >>> emulation performed by HHH1?
 >>
 >> Go back and read the part you ignored moron.
 >
 > Let the record show that Peter Olcott has neglected to identify an
 > instruction that HHH emulates differently from HHH1.
 >
 >>> Failure to provide this in your next message or within one hour of
 >>> your next post in this newsgroup will be taken as your official on-
 >>> the-record admission that the emulations performed by HHH and HHH1
 >>> are in fact exactly the same up until the point that HHH aborts, at
 >>> which point HHH did not correctly simulate the last instruction it
 >>> simulated as you are previously on record as admitting.
 >
 > Therefore, as per the above requirements:
 >
 > LET THE RECORD SHOW
 >
 > That Peter Olcott
 >
 > Has *officially* admitted
 >
 > That the emulations performed by HHH and HHH1 are in fact exactly the
 > same up until the point that HHH aborts, at which point HHH did not
 > correctly simulate the last instruction it simulated as he is previously
 > on record as admitting.