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Richard Damon <richard@damon-family.org> writes:What you are missing is that the next step of olcott is to say that when he uses the 'exact same HHH, with only some extra code to abort the simulation', it is still an infinite recursion. He does not understand that adding the abort code makes the behaviour fundamentally different. It is difficult for him to understand, because he refuses to use different names for the different versions of HHH, because he dreams that they are al exactly the same (except for small changes).On 5/8/25 7:53 PM, olcott wrote:[...]Perhaps I've missed something. I don't see anything in the above thatvoid 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;.
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?
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