Re: Every sufficiently competent C programmer knows --- Very Stupid Mistake and Liars

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