Re: DDD correctly emulated by HHH is Correctly rejected as non-halting V2

Op 15.jul.2024 om 05:35 schreef olcott:

On 7/14/2024 10:02 PM, Mike Terry wrote:

On 15/07/2024 01:20, joes wrote:

Am Sun, 14 Jul 2024 09:00:55 -0500 schrieb olcott:

On 7/14/2024 3:29 AM, joes wrote:

Am Sat, 13 Jul 2024 18:33:53 -0500 schrieb olcott:

On 7/13/2024 6:26 PM, joes wrote:

Can you elaborate? All runtime instances share the same static code.
I am talking about the inner HHH which is called by the simulated
DDD. That one is, according to you, aborted. Which is wrong, because
by virtue of running the same code, the inner HHH aborts ITS
simulation of DDD calling another HHH.

>

What are the twins and what is their difference?
Do you disagree with my tracing?

>
>

The directly executed DDD is like the first call of infinite recursion.
The emulated DDD is just like the second call of infinite recursion.
When the second call of infinite recursion is aborted then the first
call halts.

Not really. Execution does not continue.

void Infinite_Recursion()
{
    Infinite_Recursion();
}
The above *is* infinite recursion.
A program could emulate the above code and simply skip line 3 causing
Infinite_Recursion() to halt.

That would be incorrect.
>

When DDD calls HHH(DDD) HHH returns.

Therefore it does not need to be aborted.

When DDD correctly emulated by HHH the call never returns as is proven
below. The executed DDD() has HHH(DDD) skip this call.

I do not see this below.

HHH(DDD) must skip this call itself by terminating the whole DDD
process.

>

Because this HHH does not know its own machine address HHH only sees
that DDD calls a function that causes its first four steps to be
repeated. HHH does not know that this is recursive simulation. To HHH it
looks just like infinite recursion.

>

New slave_stack at:1038c4 -- create new process context for 1st DDD
Begin Local Halt Decider Simulation   Execution Trace Stored at:1138cc

>

[0000217a][001138b4][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
New slave_stack at:14e2ec -- create new process context for 2nd DDD

>

[0000217a][0015e2dc][0000217f] e853f4ffff call 000015d2 ; call HHH(DDD)
Local Halt Decider: Infinite Recursion Detected Simulation Stopped

How is this detected?

>
PO seems not to want to answer you, as I notice you've asked this question more than once and PO dodges a direct response, so I'll try. (Alternatively, PO has provided a link to his source code in the past, so if you can find that link you can just look the answer yourself - the functions are all in his halt7.c file, which is compiled but not linked, then the obj file is interpreted within his x86utm.exe (source also given in the link.  The link might not reflect his current code??)
>
Anyhow, this is what I reckon...
>
HHH [outer HHH only!] examines a global trace table of simulated instruction (from all simulation levels merged together).  The particular message "Infinite Recursion Detected Simulation Stopped" seems to be issued when:
-  last instruction is a CALL
-  working backwards through the merged trace table, another CALL is encountered
-  ..which is issued at the same address
-  ..and is calling to the same address
-  ..and no "conditional branch" instructions occur in the trace table
      between the two call instructions
>
KEY TO NOT BEING MISLED BY THE ABOVE:
>
0. The "Infinite Recursion Detected Simulation Stopped" message is just a printf.
    It does not prove that /actual/ infinite recursion was detected - on the contrary,
    all here but PO realise that the recursion detected is just finite recursion.
>
1. The trace table being examined is NOT an x86 processor trace - it is a
    "merged simulation trace" containing entries for ALL SIMULATION LEVELS.
    So the two CALL instructions are not referring to one single x86 processor.

 When emulated DDD calls HHH(DDD) the outer HHH emulates itself
emulating DDD.
 I think that joes does not understand these things.
 

    Typically, the last call instruction is from a deeper nested simulation
    than the earlier detected call instruction.  The outer simulations are all
    still running, but do not appear in the trace table or logs presented by PO
    due to the next note.
>
2. The searched trace table is filtered to only contain instructions within the C
    function D/DD/DDD/.. !!
    YES, YOU READ THAT RIGHT!  ALL CODE IN HHH IS TOTALLY IGNORED, INCLUDING
    THE CONDITIONAL BRANCH INSTRUCTIONS THAT ARE TESTING THE VERY ABORT TESTS
    THAT CAUSE OUTER HHH TO ABORT.
>
3. Inner HHH's do not perform the same tests as above, because they inspect a global
    variable which tells them they are inner HHH's.  Yeah, that means the simulation
    is completely broken logically... [but... the outer HHH will abort first, so
    PO might argue the outcome will be the same, even though logically it is
    broken...]
>
 > Is it also triggered when calling a function
 > in a loop?
>
Not sure what you mean.  Calling a function in a loop ends if the loop ends, right?  What loop are you thinking of?
>
Anyhow, provided the call instructions are physically located in function D() [i.e. not H() or something called from H] I guess it would match.  But the C function D has only one call instruction, which isn't in a loop!
>
Regards,
Mike.
>

 *I have boiled it all down to this simple tautology*
Any input that must be aborted to prevent the non termination
of simulating termination analyzer HHH necessarily specifies
non-halting behavior or it would never need to be aborted.
 

But since HHH aborts, it halts and therefore it is not involved in an infinite recursion.
So, its simulation does not need to be prevented from non termination and it is incorrect to abort its simulation.
DDD has nothing to do with it. It is easy to eliminate DDD:
        int main() {
          return HHH(main);
        }
This has the same problem. This proves that the problem is not in DDD, but in HHH, which halts when it aborts the simulation, but it decides that the simulation of itself does not halt.
HHH is unable to decide about finite recursions.
void Finite_Recursion (int N) {
   if (N > 0) Finite_Recursion (N - 1);
}
It decides after N recursions that there is an infinite recursion, which is incorrect.
Your HHH is programmed to abort the simulation after N cycles of recursive simulations. Therefore, it is incorrect to abort the simulation of HHH when the simulated HHH has performed N-1 cycles, because that changes the behaviour of HHH.
Since the simulated HHH always runs one cycle behind the simulating HHH, it is clear that HHH can never simulate enough cycles for a correct simulation, as is required by the x86 language.
Therefore, the simulation is incorrect according to the criteria you stipulated.
The conclusion is simple:
HHH cannot possibly simulate itself correctly.
No matter how much you want it to be correct, or how many times you repeat that it is correct, it does not change the fact that such a simulation is incorrect, because it is unable to reach the end.
Sipser would agree that this incorrect simulation cannot be used to detect a non-halting behaviour.