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On 5/11/2025 4:12 AM, Mikko wrote:What he says is true. However, the assumptions that HHH does a correctOn 2025-05-10 15:13:32 +0000, olcott said:On 5/8/2025 8:30 PM, Keith Thompson wrote:
On 5/10/2025 2:15 AM, Mikko wrote:You are lying again. Nothing above was told you in comp.lang.c.On 2025-05-09 03:01:40 +0000, olcott said:void DDD()
On 5/8/2025 9:23 PM, Keith Thompson wrote:What were you told in comp.lang.c that you were not told in comp.theory?Richard Damon <richard@damon-family.org> writes:*****On 5/8/25 7:53 PM, olcott wrote:[...]Perhaps I've missed something. I don't see anything in the above thatvoid DDD()And thus not correctly simulatd.
{
 HHH(DDD);
 return;
}
We don't need to look at any of my code for me
to totally prove my point. For example when
the above DDD is correctly simulated by HHH
this simulated DDD cannot possibly reach its own
"return" instruction.
Sorry, there is no "OS Exemption" to correct simulaiton;.
implies that HHH does not correctly simulate DDD. Richard, you've read
far more of olcott's posts than I have, so perhaps you can clarify.
If we assume that HHH correctly simulates DDD, then the above code is
equivalent to:
void DDD()
{
DDD();
return;
}
which is a trivial case of infinite recursion. As far as I can tell,
assuming that DDD() is actually called at some point, neither the
outer execution of DDD nor the nested (simulated) execution of DDD
can reach the return statement. Infinite recursion might either
cause a stack overflow and a probable program crash, or an unending
loop if the compiler implements tail call optimization.
I see no contradiction, just an uninteresting case of infinite
recursion, something that's well understood by anyone with a
reasonable level of programming experience. (And it has nothing to
do with the halting problem as far as I can tell, though of course
olcott has discussed the halting problem elsewhere.)
Richard, what am I missing?
Now you are seeing what I was talking about.
Now you are seeing why I needed to cross post
to comp.lang.c
{
HHH(DDD);
return;
}
People quickly realize that when DDD is correctly
simulated by HHH that DDD cannot possibly reach
its "return" statement (final halt state).
Once you know this then you can see that the
same thing applies to DD.
int DD()
{
int Halt_Status = HHH(DD);
if (Halt_Status)
HERE: goto HERE;
return Halt_Status;
}
Once you know this then you know that the halting
problem's otherwise "impossible" input is non-halting.
Once you know this then you know that the halting
problem proof has been correctly refuted.
> Assuming that HHH(DDD) "correctly simulates" DDD, and assuming it
> does nothing else, your code would be equivalent to this:
>
> void DDD(void) {
> DDD();
> return;
> }
>
> Then the return statement (which is unnecessary anyway) will never be
> reached. In practice, the program will likely crash due to a stack
> overflow, unless the compiler implements tail-call optimization, in
> which case the program might just run forever -- which also means the
> unnecessary return statement will never be reached.
>
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