Re: Every sufficiently competent C programmer knows --- Semantic Property of Finite String

On 3/12/2025 6:31 PM, olcott wrote:

On 3/12/2025 5:03 PM, dbush wrote:

On 3/12/2025 5:38 PM, olcott wrote:

On 3/12/2025 3:53 PM, dbush wrote:

On 3/12/2025 4:29 PM, olcott wrote:

On 3/12/2025 2:16 PM, dbush wrote:

On 3/11/2025 10:46 PM, olcott wrote:

On 3/11/2025 9:41 PM, dbush wrote:

On 3/11/2025 10:39 PM, olcott wrote:

On 3/11/2025 9:37 PM, dbush wrote:

On 3/11/2025 10:36 PM, olcott wrote:

On 3/11/2025 9:32 PM, dbush wrote:

On 3/11/2025 10:31 PM, olcott wrote:

On 3/11/2025 9:18 PM, dbush wrote:

On 3/11/2025 10:06 PM, olcott wrote:

On 3/11/2025 9:02 PM, dbush wrote:

On 3/11/2025 9:41 PM, Richard Heathfield wrote:

On 12/03/2025 01:22, olcott wrote:

DDD correctly simulated by HHH never reaches its
own "return" instruction and terminates normally
in any finite or infinite number of correctly
simulated steps.

>
If it correctly simulates infinitely many steps, it doesn't terminate. Look up "infinite".
>
But your task is to decide for /any/ program, not just DDD. That, as you are so fond of saying, is 'stipulated', and you can't get out of it. The whole point of the Entscheidungsproblem is its universality. Ignore that, and you have nothing.
>

>
>
Given that his code has HHH(DD) returning 0,

>
THESE ARE THE WORDS ANYONE THAT DODGES THESE
WORDS WILL BE TAKEN FOR A LIAR
>
void DDD()
{
   HHH(DDD);
   return;
}
>
DDD correctly simulated by HHH never reaches its
own "return" instruction and terminates normally
in any finite or infinite number of correctly
simulated steps.
>

>
Changing the input is not allowed.

>
*You are simply lying that any input was ever changed*
>

>
You did precisely that when you hypothesize different code for HHH.
>
Changing the input is not allowed.

>
*THIS IS WHAT MY ORIGINAL WORDS MEANT*
HHH is the infinite set of every possible C function
that correctly emulates N steps of its input where
N any finite positive integer.
>

>
In other words, you're changing the input.
>
Changing the input is not allowed.

>
It is an infinite set of HHH/DDD pairs having the
property that DDD[0] ... DDD[N] never halts.
>

>
In other words, you're not answering the question that a solution to the halting problem is required to answer:
>
(<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
>

>
Yes I am yet you refuse to pay anywhere near close
enough attention to see how I already fully addressed this.
If you pay 100% perfect attention you might get it.
>

>
False.  (<DDD>,null) maps to 1 as per the above requirements, but your HHH maps (<DDD>,null) to 0, therefore it fails to meet the requirements.

>
<unrelated copy-paste response>
>

>
>
So no response?  I'll take it that you agree with the above.

>
Making sure to always give credit where credit is due this
point in our conversation is the point where I first translated
my perspective into the semantic property of a finite string.
>
A decider is required to report on a semantic (or syntactic)
property of its input finite string (even if Rice incorrectly
says this is impossible in this case) and not allowed to report
on any damn thing else.
>
The fact that DDD calls HHH(DDD) in recursive emulation
<is> an aspect of the semantics of the input finite string
that cannot be correctly ignored.
>
>

>
Remember the stipulative definition of a solution to the halting problem:
>
>

 Is to map the input finite string to the semantic property
of this finite string. Any other mapping contradicts the
definition of a decider.

And that property is as follows:
(<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
I should also point out that I never mentioned anything about a "decider", simply "a solution to the halting problem".  Neither did Linz.