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

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Sujet : Re: DDD specifies recursive emulation to HHH and halting to HHH1
De : noreply (at) *nospam* example.org (joes)
Groupes : comp.theory
Date : 29. Mar 2025, 10:19:09
Autres entêtes
Organisation : i2pn2 (i2pn.org)
Message-ID : <c8d42d8ae414c4a6aba7d6f3ccc7892ee728edc7@i2pn2.org>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
User-Agent : Pan/0.145 (Duplicitous mercenary valetism; d7e168a git.gnome.org/pan2)
Am Fri, 28 Mar 2025 17:38:22 -0500 schrieb olcott:
On 3/28/2025 5:30 PM, dbush wrote:
On 3/28/2025 6:09 PM, olcott wrote:
On 3/28/2025 3:38 PM, dbush wrote:
On 3/28/2025 4:30 PM, olcott wrote:
On 3/28/2025 2:24 PM, dbush wrote:
On 3/28/2025 3:21 PM, olcott wrote:
On 3/28/2025 4:43 AM, Fred. Zwarts wrote:
Op 28.mrt.2025 om 03:13 schreef olcott:
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:

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.
>
If A TM can only compute the mapping from *its* input to *its*
output, it cannot be wrong.
>
Taking a wild guess does not count as computing the mapping.
>
False.  The only requirement is to map a member of the input domain
to a member of the output domain as per the requirements.
If it does so in all cases, the mapping is computed.  It doesn't
matter how it's done.
>
Unless an input is transformed into an output on the basis of a
syntactic or semantic property of this input it is not a Turing
computable function.
int StringLength(char *S)
{
   return 5;
}
Does not compute the string length of any string.
>
False.  It computes the length of all strings of length 5.
>
It does not compute (a sequence of steps of an algorithm that derive
an output on the basis of an input) jack shit it makes a guess.
Even a constant function is a "computation", even if it doesn't actually
do any work.

Doesn't matter. If the requirement is to return 5 for strings that have
a length of 5, it meets the requirement.
 
The actual requirement is to compute the mapping from a finite string to
its length using a sequence of algorithmic steps.
Likewise for halting. Compute the mapping from a finite string of
machine code to the behavior that this finite string specifies.
Do you reckon the direct execution of a TM contradicts the specification?

--
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.

Date Sujet#  Auteur
5 Jan 26 o 

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