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On 5/9/2025 7:23 PM, Mike Terry wrote:It will certainly help. Not in making a better simulator, but by creating an input DDD that halts. (But your HHH misses that verifiable fact.)On 09/05/2025 03:23, Keith Thompson wrote:That was not in the specification that he responded to.Richard Damon <richard@damon-family.org> writes:>On 5/8/25 7:53 PM, olcott wrote:[...]>void DDD()>
{
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.
And thus not correctly simulatd.
>
Sorry, there is no "OS Exemption" to correct simulaiton;.
Perhaps I've missed something. I don't see anything in the above that
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?
>
Depends on what you've picked up on.
>
Do you get that HHH's simulation is a /partial/ simulation? HHH is free
Thus within this specification DDD correctly simulated
by HHH cannot possibly reach its final halt state in an
infinite number of steps.
If an infinite number steps steps is insufficient
then how can less an infinite steps possibly help?
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