Re: Simulating termination analyzers by dummies --- What does halting mean?

On 2024-06-21 13:19:28 +0000, olcott said:

On 6/21/2024 2:11 AM, Mikko wrote:

On 2024-06-20 15:23:09 +0000, olcott said:
 

On 6/20/2024 10:08 AM, Mikko wrote:

On 2024-06-20 05:40:28 +0000, olcott said:
 

On 6/20/2024 12:29 AM, Mikko wrote:

On 2024-06-19 14:05:29 +0000, olcott said:
 

On 6/19/2024 4:29 AM, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:

On 6/18/2024 4:36 PM, Alan Mackenzie wrote:

[ Followup-To: set ]

 

In comp.theory olcott <polcott333@gmail.com> wrote:

On 6/18/2024 12:57 PM, joes wrote:

Am Tue, 18 Jun 2024 12:25:44 -0500 schrieb olcott:

On 6/18/2024 12:06 PM, joes wrote:
void DDD()
{
H0(DDD);
}
DDD correctly simulated by any H0 cannot possibly halt.

DDD halts iff H0 halts.

 

So H0 returns "doesn't halt" to DDD, which then stops running,
so H0 should have returned "halts".

 

This was three messages ago.
I had to make sure that you understood that halting
does not mean stopping for any reason and only includes
the equivalent of terminating normally.

 

No.  You're wrong, here.  A turing machine is either running or it's
halted.  There's no third alternative.  If your C programs are not in one
of these two states, they're not equivalent to turing machines.

 

Although I agree with this there seems to be nuances of
disagreement across the experts.

 I doubt that very much.  The whole point of turing machines is to remove
ambiguity and unneeded features from the theory of computation.  A third
alternative state is unneeded.
 

 Some people say that a TM can halt in a non-final state.

 People may use different words to express the same facts. What some
people call "halting in a non-final state" is called "rejecting" by
some other people. But the facts are what they are independently of
the words used to express them.

 Ambiguity and vagueness make communication less effective.

 As does use of common words and expressions for uncommon meanings.
 

I use C because there are zero gaps in exactly what it means.

 THere are lont of gaps in C. Some are mistakes that are corrected in
technical corrigenda. Others are undefined and implementation defined
behaviour. Your program uses non-standard extensions to C so it does
not communicate well. If also is too big to be a part of a publishable
article.
 

 *There are zero gaps in the behavior of DDD correctly simulated by HH0*
https://liarparadox.org/HH0_(DDD)_Full_Trace.pdf
 _DDD()
[00002093] 55               push ebp
[00002094] 8bec             mov ebp,esp
[00002096] 6893200000       push 00002093 ; push DDD
[0000209b] e853f4ffff       call 000014f3 ; call HH0
[000020a0] 83c404           add esp,+04
[000020a3] 5d               pop ebp
[000020a4] c3               ret
Size in bytes:(0018) [000020a4]
 Whereas the Linz specification of Ĥ says that embedded_H
does something or other that is totally unspecified:
 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn

 Linz Ĥ is fully defined in terms of H, so its behaviour can be inferred
from the behaviour of H. Therefore Linz can prove about the behaviour of
both Ĥ and H what needs be proven.

 (a) Ĥ copies its input ⟨Ĥ⟩
(b) Ĥ invokes embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(c) embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(d) simulated ⟨Ĥ⟩ copies its input ⟨Ĥ⟩
(e) simulated ⟨Ĥ⟩ invokes simulated embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩
(f) simulated embedded_H simulates ⟨Ĥ⟩ ⟨Ĥ⟩
(g) goto (d) with one more level of simulation

Linz says nothing about simulations or other methods H may use. Those are already fully determined by the construction. But if H does those (c)..(g)
then so does Ĥ.

Two complete simulations show a pair of identical TMD's are simulating a pair of identical inputs.  We can see this thus proving recursive simulation.

Yes. However, this observation is not used in steps (c)..(g).
--
Mikko