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

On 5/8/25 10:22 PM, olcott wrote:

On 5/8/2025 9:02 PM, Keith Thompson wrote:

olcott <polcott333@gmail.com> writes:

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.

>
You see, this is something you've gotten wrong, and you need somebody
who can explain that to you in terms you can understand and accept.
>

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.

>
Is there any reason you couldn't have written that as follows?
>
void DDD(void)
{
   HHH(DDD);
}
>
You could then talk about it not reaching its closing brace rather
than not reaching its "return" instruction.  BTW, it's correctly
called a "return statement" in C; dropping it would make it easier
to avoid your incorrect use of terminology.  (Assembly or machine
code has "instructions"; C has "statements" and "declarations".)
>

DDD correctly simulated by HHH is the same thing
as infinite recursion between HHH and DDD yet is
implemented as recursive simulation.

>
Sure, infinite recursion is infinite, regardless of how it's
implemented, assuming it's implemented correctly.  That's so trivally
obvious that I simply don't believe that "the comp.sci" people are
failing to understand it -- though I can believe that you believe it.
>

 Look at their replies to this post.
Not a one of them will agree that
 void DDD()
{
   HHH(DDD);
   return; // final halt state
}
 When 1 or more instructions of DDD are correctly
simulated by HHH then the correctly simulated DDD cannot
possibly reach its "return" instruction (final halt state).

Because that isn't the definiton of Non-Halting.
And, your emulators are violating the rules they need to follow by looking at memory that isn't part of their input.

 They have consistently disagreed with this
simple point for three years.

because your statement is irrelvent, and a violation of the criteria you aggreed you needed to follow.
If you can't follow your own rules, you are just being a liar.

 

I doubt that any such person exists, but only for reasons related
to you.
>

>