Re: Halting Problem: What Constitutes Pathological Input

Am Wed, 07 May 2025 22:32:12 -0500 schrieb olcott:

On 5/7/2025 10:00 PM, Mike Terry wrote:

On 08/05/2025 00:01, olcott wrote:

On 5/7/2025 5:06 PM, Mike Terry wrote:

On 07/05/2025 22:32, olcott wrote:

On 5/7/2025 4:20 PM, Mike Terry wrote:

On 07/05/2025 17:25, olcott wrote:

On 5/7/2025 10:44 AM, Mike Terry wrote:

You need to read what posters actually say.  I said the traces
were the same up to the point where HHH stops simulating.

>
THAT IS COUNTER-FACTUAL.
HHH continues to emulate DD long after their paths diverge.
HHH1 only emulates DD once.
HHH emulates DD once and then emulates itself emulating DD.

Where do they diverge? How does HHH1 simulate the HHH called from DD
returning?

 I didn't say anything about calls that return or do not return
"being the same thing" and none of what you relates to whether
what I said was correct.
>
Look, if you insist the traces are not the same up to the point
where HHH stops simulating, show the two traces and we'll just
look and see! Simples.
>

HHH1(DD) emulates DD once.
HHH(DD) emulates DD once and then emulates itself emulating DD.

Excellent - above we have the trace for HHH1, half of what we need.
While we /could/ use that to /deduce/ what the trace for HHH should
be, we shouldn't have to resort to that.  The clean way to proceed
is for you to now post the similar trace for main calling HHH, then
we can compare them with minimal editing...
>

The two calls to HHH(DD) are in the part you ignored.
HHH1(DD) emulates DD that calls the emulated HHH(DD) to emulate DD
that calls HHH(DD) that emulates itself emulating DD.
I marked these calls with comments.

>
As I explained, it would be clearer if you just supply the requested
trace.
Are you unable/unwilling to do that?
>

[identical traces]

>
Thanks for that.
Can you see from the very traces you've posted that what I and DBush
earlier said is exactly right? That makes nonsense of your claims that
what was said was "counter-factual".
It's all there in the traces.  I'll be posting full details in a new
thread...
 

It is NOT true that the first difference in the behavior is when HHH
aborts its emulation of DD.
The difference in the behavior begins long before that. It begins as
soon as the the DD emulated by HHH1 calls HHH(DD) and the DD emulated by
HHH calls HHH(DD).
 
The first of these calls DOES return.
as proven by the execution trace.
The second of these calls cannot possibly return.
as proven by repeating sequence in the execution trace.
 
HHH1 never emulates itself emulating DD because the call from DD to
HHH(DD) returns.
HHH continues to emulate itself emulating DD proving a different
sequence than DD emulated by HHH1. The call from the emulated DD to
HHH(DD) cannot possibly return.
 
This is really only a variation of the same idea as infinite recursion
between two functions. The only difference here is that one of these
functions has the power to stop it.
This causes the other function to stop running yet not halt because
halting is reserved for reaching a final halt state.

Um, those two traces are the same. I thought HHH1 didn't abort?
If the difference were that one call does not return, something must
happen before. It can't be an abort, because an aborted simulation
does not return. How do HHH and HHH1 simulate DD calling HHH
differently?
HHH1 obviously simulates HHH simulating DD, because HHH is what DD calls.
How does it simulate HHH returning?


--
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.