Re: Why I need to cross-post to comp.lang.c --- CORRECTLY REFUTED

On 2025-05-12 17:19:52 +0000, olcott said:

On 5/12/2025 12:08 PM, dbush wrote:

On 5/12/2025 12:59 PM, olcott wrote:

On 5/12/2025 11:42 AM, dbush wrote:

On 5/12/2025 12:24 PM, olcott wrote:

On 5/12/2025 9:53 AM, dbush wrote:

On 5/12/2025 10:47 AM, olcott wrote:

On 5/12/2025 2:45 AM, Mikko wrote:

On 2025-05-11 16:03:29 +0000, olcott said:
 

On 5/11/2025 4:12 AM, Mikko wrote:

On 2025-05-10 15:13:32 +0000, olcott said:
 

On 5/10/2025 2:15 AM, Mikko wrote:

On 2025-05-09 03:01:40 +0000, olcott said:
 

On 5/8/2025 9:23 PM, Keith Thompson wrote:

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?
 

*****
Now you are seeing what I was talking about.
Now you are seeing why I needed to cross post
to comp.lang.c

 What were you told in comp.lang.c that you were not told in comp.theory?

 void DDD()
{
   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.

 You are lying again. Nothing above was told you in comp.lang.c.

 On 5/8/2025 8:30 PM, Keith Thompson wrote:
 > 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.
 >

 What he says is true. However, the assumptions that HHH does a correct
simulation and that it does nothing else are not.
 

 _DDD()
[00002172] 55         push ebp      ; housekeeping
[00002173] 8bec       mov ebp,esp   ; housekeeping
[00002175] 6872210000 push 00002172 ; push DDD
[0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
[0000217f] 83c404     add esp,+04
[00002182] 5d         pop ebp
[00002183] c3         ret
Size in bytes:(0018) [00002183]
 We make an infinite set of purely hypothetical HHH pure
x86 emulators each one is at machine address 000015d2
thus called by DDD.

 And each DDD is a distinct algorithm.
 

 Such that none of them halt.
The only difference is the number of steps of DDD
emulated by HHH.

 And the halt status of algorithm DDD has nothing to do with the halt status of algorithm DDD1, DDD2, etc.
 

 When the only difference between HHH/DDD pairs is that

 They are different algorithms and therefore the halt status of one has nothing to do with the halt status of another.

 That is not true, this difference is not
the kind that can possibly change the halt status.
 The difference is merely applying mathematical
induction to the exact same algorithm of HHH
emulating N steps of DDD.

It is not the exact same algorithm if the number of the emulated steps
is not the same. N is not an input HHH but a part of or a consequence
of the algorthm.
--
Mikko