Re: Incorrect requirements --- Computing the mapping from the input to HHH(DD)

On 5/9/2025 4:00 AM, Fred. Zwarts wrote:

Op 09.mei.2025 om 02:05 schreef olcott:

On 5/8/2025 6:54 PM, Keith Thompson wrote:

olcott <polcott333@gmail.com> writes:

On 5/8/2025 6:30 PM, Richard Heathfield wrote:

On 08/05/2025 23:50, olcott wrote:

[...]

If you are a competent C programmer

Keith Thompson is a highly-respected and very competent C
programmer.

>
*Then he is just who I need*

>
No, what you need is someone who is an expert in mathematical logic
(I am not) who can explain to you, in terms you can understand and
accept, where you've gone wrong.  Some expertise in C could also
be helpful.
>

>
The key gap in my proof is that none of the comp.sci
people seems to have a slight clue about simple C
programming.
>
void DDD()
{
   HHH(DDD);
   return;
}
>
*THIS IS THE C PART THAT NO ONE HERE UNDERSTANDS*
DDD correctly simulated by HHH cannot possibly
reach its own "return" instruction.

 The HHH that you propose and is included in DDD does halt. But it is known that this HHH prematurely aborts which makes it impossible for the simulation to reach the reachable 'return'. HHH simply ignores the conditional abort in the simulation.
 

void DDD()
{
   HHH(DDD);
   return;
}
When 1 or more statements of DDD are correctly
simulated by HHH then this correctly simulated
DDD cannot possibly reach its own “return statement”.
--
Copyright 2024 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer