Re: Simulating termination analyzers by dummies --- criteria is met

On 2024-06-22 14:11:28 +0000, olcott said:

On 6/22/2024 8:27 AM, Richard Damon wrote:

On 6/22/24 9:04 AM, olcott wrote:

On 6/22/2024 3:05 AM, Mikko wrote:

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

 I am the sole inventor of the simulating halt decider.
 Ben Bacarisse contacted professor Sipser to verify that he
really did says this. The details are in this forum about
the same date.
 https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X/   <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
   If simulating halt decider H correctly simulates its input D
   until H correctly determines that its simulated D would never
   stop running unless aborted then
    H can abort its simulation of D and correctly report that D
   specifies a non-halting sequence of configurations.
</MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>

 And, as I remember, he also verified that he disagrees with your definition of correct simulation.
 

 *Ben also verified that the criteria have been met*
On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game. PO really /has/ an H
 > (it's trivial to do for this one case) that correctly determines
 > that P(P) *would* never stop running *unless* aborted.

 Right, Ben was willing to do what I am not that you can prove that, by your definition, H can show that it "must" abort its simulation or the input will run forever.
 But, just like me, he also agrees that this is NOT the defintion of Halting, so H is just shown to be a correct (partial) POOP decider but ot a Halt Decider, not even for that one input.
 

 On 10/14/2022 7:44 PM, Ben Bacarisse wrote:
 > I don't think that is the shell game. PO really /has/ an H
 > (it's trivial to do for this one case) that correctly determines
 > that P(P) *would* never stop running *unless* aborted.
 >
 > He knows and accepts that P(P) actually does stop. The
 > wrong answer is justified by what would happen if H
 > (and hence a different P) where not what they actually are.
 >
*Ben agrees that the criteria is met for the input*
 Computable functions are the formalized analogue of the
intuitive notion of algorithms, in the sense that a function
is computable if there exists an algorithm that can do the
job of the function, i.e. *given an input of the function*
*domain it can return the corresponding output*
https://en.wikipedia.org/wiki/Computable_function
 *Ben disagrees that the criteria is met for the non-input*
Yet no one here can stay focused on the fact that non-inputs
*DO NOT COUNT*

In particular, you can't. You have insisted that your "decider"
or "anlyzer" (or whatever word you happen to use) H or HH (or
hwatever name you happen to use) must return false because a
non-input (where instead of the actually called function another
function that does not halt is called) does not halt.

void DDD()
{
   HHH0(DDD);
}
 int main()
{
   Output("Input_Halts = ", HHH0(DDD));
   Output("Input_Halts = ", HHH1(DDD));
}
 It is a verified fact that the behavior that finite string DDD
presents to HH0 is that when DDD correctly simulated by HH0
calls HH0(DDD) that this call DOES NOT RETURN.
 It is a verified fact that the behavior that finite string DDD
presents to HH1 is that when DDD correctly simulated by HH1
calls HH0(DDD) that this call DOES RETURN.

--
Mikko