Re: DDD specifies recursive emulation to HHH and halting to HHH1

On 3/27/2025 9:04 PM, Richard Damon wrote:

On 3/27/25 9:07 PM, olcott wrote:

On 3/27/2025 7:38 PM, dbush wrote:

On 3/27/2025 8:34 PM, olcott wrote:

On 3/27/2025 7:12 PM, dbush wrote:

On 3/27/2025 8:11 PM, olcott wrote:

On 3/27/2025 7:02 PM, dbush wrote:

On 3/27/2025 7:36 PM, olcott wrote:

On 3/27/2025 1:27 PM, dbush wrote:

On 3/27/2025 1:50 PM, olcott wrote:

On 3/27/2025 2:18 AM, Fred. Zwarts wrote:

Op 27.mrt.2025 om 04:09 schreef olcott:

On 3/26/2025 8:22 PM, Richard Damon wrote:
>
_DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov  ebp,esp  ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add  esp,+04
[00002182] 5d         pop  ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
>

Non-Halting is that the machine won't reach its final staste even if an unbounded number of steps are emulated. Since HHH doesn't do that, it isn't showing non-halting.
>

>
DDD emulated by any HHH will never reach its final state
in an unbounded number of steps.
>
DDD emulated by HHH1 reaches its final state in a finite
number of steps.
>

It is not very interesting to know whether a simulator reports that it is unable to reach the end of the simulation of a program that halts in direct execution.

>
That IS NOT what HHH is reporting.
HHH correctly rejects DDD because DDD correctly
emulated by HHH cannot possibly reach its own
final halt state.

>
In other words, HHH is not a halt decider because it is not computing the required mapping:
>

>
Troll
>

>
On Monday, March 6, 2023 at 3:19:42 PM UTC-5, olcott wrote:
 > In other words you could find any error in my post so you resort to the
 > lame tactic of ad hominem personal attack.
>

>
Troll
>

>
On 7/22/2024 10:51 AM, olcott wrote:
 > *Ad Hominem attacks are the first resort of clueless wonders*
>

>
I corrected your error dozens of times and you
ignore these corrections and mindlessly repeat
your error like a bot

>
Which is what you've been doing for the last three years.
>
Projection, as always.  I'll add the above to the list.
>

>
TM's cannot possibly ever report on the behavior
of the direct execution of another TM. I proved
this many times in may ways. Ignoring these proofs
IT NOT ANY FORM OF REBUTTAL.
>

 Sure they can.
 WHere is your proof? And what actual accepted principles is is based on?
 

No TM can take another directly executed TM as an input
and Turing computable functions only compute the mapping
from inputs to outputs.
--
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer