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

On 3/12/2025 4:32 AM, Fred. Zwarts wrote:

Op 12.mrt.2025 om 03:39 schreef olcott:

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,

>
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.

>
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.
>

Proving that HHH[0] ... HHH[N} are unable to correctly complete the
simulation.

 
void Infinite_Loop()

void Infinite_Recursion()

In the exact same way that HHH cannot complete the simulation of the
above functions.
BECAUSE THEY SPECIFY NON-TERMINATING BEHAVIOR.