Re: HHH(DDD) does not see the exact same behavior pattern as HHH(Infinite_Recursion) OLCOTTS ERROR

On 7/29/24 8:16 PM, olcott wrote:

On 7/29/2024 5:57 PM, Mike Terry wrote:

On 29/07/2024 20:36, Fred. Zwarts wrote:

Op 29.jul.2024 om 15:03 schreef olcott:

On 7/29/2024 2:29 AM, Fred. Zwarts wrote:

Op 28.jul.2024 om 22:10 schreef olcott:

On 7/28/2024 2:14 PM, Fred. Zwarts wrote:

Op 28.jul.2024 om 16:25 schreef olcott:

On 7/28/2024 2:59 AM, Fred. Zwarts wrote:

Op 28.jul.2024 om 05:15 schreef olcott:

On 7/27/2024 7:40 PM, Mike Terry wrote:

On 27/07/2024 19:14, Alan Mackenzie wrote:

olcott <polcott333@gmail.com> wrote:
>

Stopping running is not the same as halting.
DDD emulated by HHH stops running when its emulation has been aborted.
This is not the same as reaching its ret instruction and terminating
normally (AKA halting).

>
I think you're wrong, here.  All your C programs are a stand in for
turing machines.  A turing machine is either running or halted. There is
no third state "aborted".  An aborted C program certainly doesn't
correspond with a running turing machine - so it must be a halted turing
machine.
>
So aborted programs are halted programs.  If you disagree, perhaps you
could point out where in my arguments above I'm wrong.
>

>
Aborting is an action performed by a simulator, not the TM being simulated.
>
If we want to count "aborted" as some kind of final status, it will have to be the status of a specific /PARTIAL SIMULATOR's/ simulation of a given computation.  That's not a property of the computation itself, as different partial simulators can simulate the same computation and terminate for different reasons.  Like HHH(DDD) aborts, while UTM(DDD) simulates to completion and so the final simulation status is halts. [Neither of those outcomes contradict the fact that the computation DDD() halts.]
>
If some partial simulator halts when simulating a computation [as with UTM(DDD)] that implies the computation DDD() halts. But if the simulator aborts, it doesn't say that much (in and of itself) about whether the /computation/ halts.  The halting problem statement is not concerned with simulations or how the simulations end.
>
Every time anyone in these PO threads says "halts" it ought to be 100% clear to everyone whether the computation itself is being discussed, or whether some simulation final status is intended. (But that's far from being the case!)  Since the halting problem is concerned with computations halting and not how partial simulations are ended, I suggest that PO explicitly make clear that he is referring to simulations, whenever that is the case. It seems reasonable that readers seeing "halts" with no further clarification can interpret that as actual computation behaviour, as that is how the term is always used in the literature.  Same with other terms like "reach"...
>
So when PO says "DDD simulated by HHH cannot reach its final ret instruction" is he talking about the computation DDD() [as defined mathematically], or its simulation by HHH?  He means the latter, but its far from clear, I'd say!  [I think most readers now have come around to reading it as a statement about simulations rather than the actual computation, which totally changes things...]
>
>
Mike.
>

>
_DDD()
[00002163] 55         push ebp      ; housekeeping
[00002164] 8bec       mov ebp,esp   ; housekeeping
[00002166] 6863210000 push 00002163 ; push DDD
[0000216b] e853f4ffff call 000015c3 ; call HHH(DDD)
[00002170] 83c404     add esp,+04
[00002173] 5d         pop ebp
[00002174] c3         ret
Size in bytes:(0018) [00002174]
>
It is a verified fact that DDD emulated by HHH 100% exactly
and precisely according to the actual x86 semantics of
the emulated code including the recursive call that causes
HHH to emulate itself emulating DDD cannot possibly get
past it own 0000216b machine address.
>
*Anyone as much as hinting otherwise is not being truthful*
If we remove all of the problematic code then this same
trace still occurs until it crashes from OOM error.
>

>
No matter how much olcott wants it to be correct, or how many times olcott repeats that it is correct, it does not change the fact that such a simulation is incorrect, because it is unable to reach the end.

>
It is ridiculously stupid to expect the correct emulation
of a non-halting input to end.

>
Irrelevant nonsense ignored.
We are not discussing a non-halting HHH,

>
No we are not. Please don't act so stupidly.

>
Why are you contradicting yourself so often? It does not look very clever.
You were talking about an infinite set of HHH, some of which abort, some of which do not abort.

>
*This is all that I am willing to talk about with you*
Every change of subject away from this point will be
construed as the dishonest dodge of the strawman deception.
>
You didn't even bother to look at how HHH examines the
execution trace of Infinite_Recursion() to determine that Infinite_Recursion() specifies non-halting behavior.
>
Because of this you cannot see that the execution trace
of DDD correctly emulated by DDD is essentially this same
trace and thus also specifies non-halting behavior.

>
That is only because you are cheating, by hiding the conditional branch instructions of HHH, which should follow the call instruction into HHH.
HHH simulating itself is more like
>
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
>
You never bothered to think about it.

>
Also there is the crucial difference that Infinite_Recursion() trace is a trace for a single x86 processor.  The HHH/DDD trace is not a single processor trace, as it contains entries for multiple virtual x86 processors, all merged into one.  There are all sorts of argument that can be applied to the simple single x86 processor trace scenario, that simply don't work when transferred to a multi-processor-simulation merged trace.  PO doesn't understand these differences, and has said there is NO difference!  He also deliberately tries to hide these difference, by making his trace output resemble a single-processor trace as far as he can:
>

 The simple fact that you continue to ignore is that DDD
is correctly emulated by DDD according to the semantics
of the x86 instructions of DDD and HHH that includes that
DDD does call HHH(DDD) in recursive emulation that will
never stop running unless aborted.

How do you say that? The call HHH instruction is *NOT* correctly emulated.
You ignore this error to just prove you are a pathological liar.
Please show where in the Intel documentation there is support for your claim.
Your failure to do this has PROVED that you are just a pathetic liar

 

-  suppressing trace entries in H which would make it obvious that the matching calls

 I am not suppressing any freaking trace entries
https://liarparadox.org/HHH(DDD)_Full_Trace.pdf

Which, as has been pointed out before, isn't a listing of the trace that any HHH made, so irrelvent, except to prove that you are just a liar.

 It is exactly as I have been saying all along
if you can freaking understand a 1.5 page trace
then adding a 200 more pages cannot possibly help.

But both the 1.5 page trace, and the 200 page trace are wrong as the trace of the x86 emulation that HHH does.

 When Ĥ is applied to ⟨Ĥ⟩
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
Ĥ.q0 ⟨Ĥ⟩ ⊢* embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
 (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

 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.

But not INFINITE, unless embedded_H NEVER aborts, and thus not a decider.
If embedded_H ever aborts, then the simulatation started at the first instance of (c) will abort (removing all the following layers) and then the machine will halt.

 *For context here is the full Linz proof*
https://www.liarparadox.org/Linz_Proof.pdf
 There aren't any freaking virtual x86 processors there.
There are still simulation levels.
It doesn't freaking make any damn difference.

Nope, no "simulation levels" as those don't actually exist, but are just a meta-logic. There is the operation of the machine H (H^) (H^) and the machine H^ (H^).
If H (H^) (H^) goes to qn to say its input is not halting, then the copy of it in H^ (H^) does the same, and H^ (H^) Halts, showing H was just wrong.

 

    to HHH are from completely separate logical x86 processors.  The output as he presents it
    looks pretty much identical to the corresponding (but totally different character) CALL
    scenario where HHH calls DDD rather than simulating it.
>
-  omitting a "simulation level" column in his output.  He included that info for a brief
    period, following my suggestion, but then removed it, saying it "confused" readers.
    Of course it did no such thing, but PO thinks anyone who disagrees with him is
    confused (or stupid or ignorant).  So perhaps the column actually helped people understand
    or at least question what the trace was actually saying, and PO didn't like that.)
>
>
>
Mike.
>