Re: How Partial Simulations correctly determine non-halting ---Should I quit Richard at this point?

On 6/8/24 11:32 AM, olcott wrote:

On 6/8/2024 10:15 AM, Richard Damon wrote:

On 6/8/24 11:07 AM, olcott wrote:

On 6/8/2024 9:54 AM, Richard Damon wrote:

On 6/8/24 10:20 AM, olcott wrote:

On 6/8/2024 9:10 AM, Richard Damon wrote:

>
I HAVE pointed out what is missing, ANY set of truth-perserving operations from the accepted facts (which will of course need to name the fact they are working from) to your conclusion.

>
The accepted facts are here
(a) The x86 language
(b) The notion of an x86 emulator
>
{The proof that No DDD correctly emulated by any x86
  emulator H can possibly reach its own [00001df6] instruction}

>
So, how do you show this claim?
>
Do you have a tracing of the full INFINITE SET of possible Hs?
>

>
Is the set of possible execution traces of DDD correctly
emulated by x86 emulator HH on the basis of the above
accepted facts.
>
Maybe you are just clueless about these technical details
are are trying to hide this with pure bluster.
>
_DDD()
[00001de2] 55         push ebp
[00001de3] 8bec       mov ebp,esp
[00001de5] 8b4508     mov eax,[ebp+08]
[00001de8] 50         push eax         ; push DD
[00001de9] 8b4d08     mov ecx,[ebp+08]
[00001dec] 51         push ecx         ; push DD
[00001ded] e890f5ffff call 00001382    ; call HH
[00001df2] 83c408     add esp,+08
[00001df5] 5d         pop ebp
[00001df6] c3         ret
Size in bytes:(0021) [00001df6]
>
You keep disagreeing with the fact that DDD correctly
emulated by x86 emulator HH only has one single correct
execution trace of repeating the fist seven lines until
out-of-memory error.
>

>
But that is an INCORRECT trace per your definition,
>
The call HH instruction MUST be simulated into HH because that IS the behavior of the x86 instruction.

>
Did I ever say that it is not?
For the above DDD correctly emulated by x86 emulator HH
the first seven instructions of DD keep repeating because
DDD keeps calling HH(DDD,DDD) to emulate itself again and
again until HH/DDD hits out-of-memory exception.

>
So the x86 emulation of the code must go into HH(DDD,DDD)
>

 It is pretty stupid to assume otherwise when HH is
stipulated to be an x86 emulator.

Right, so why did you say otherwise?

 

The correct x86 emulation of the call to HH(DDD,DDD) will NEVER get to the sequence of instrucitions starting at 00001DE2, as the code will never jump there to just execute it.
>

 Your are saying that incorrectly DDD correctly emulated by
x86 emulator HH cannot possibly reach it own machine address
of [00001df6].

I said nothing about that. You are just serving Herring with Red Sauce agian.
The CORRECT simulation of DDD can NEVER get back to the sequence of instructions at 00001DE2, as there is never a jump to that address, only the emulator starting an emuation of that address, and the correst simulation of a simulatore is NOT the code the simulated simulator is looking at, but the code of the simulator doing the simulation.

 

By your code, the simulator will "Debug Step" those instructions.
>

 The underlying details of one HH are irrelevant when I reference
an infinite set:\

But not for your definition of the simulation.

 {The proof that No DDD correctly emulated by
any x86 emulator H
any x86 emulator H
any x86 emulator H
any x86 emulator H
any x86 emulator H
any x86 emulator H
can possibly reach its own [00001df6] instruction}

It seems that by your current analysis, it can't get past the instruction at 00001DED as there is nothing to simulate after that.
Remember, your definition said to simulate the instructions in the strict order they were reached.
If we don't have the instruction at 00001382, the simulation has to stop, as we can't go on.

 

By a pure emulator, that would mean translating the machine code into the operations it will perform, and then manipulating the virtual register set being kept by the emulator.
>

 libx86emu does that.

And if HH is a pure emulator, it need to do it to (or let libx86emu do it for it). and HH can't interfear with that process by not following each instruction with the instruction that follows it.
So, does the libx86emu keep a seperate logging of the instructions that HH is emulating, as would be needed for HH to be able to examine that trace.
The previous trace you posted wasn't the simulation that HH was doing, but was a trace of the execution of HH iteself.
It seems that HH doesn't actual have a trace of what it did available (or at least you didn't show it).
But then, you always seemed to have gotten the "levels" of simulation incorrect.

 

If your "simulation" is ACTUALLY being done using the debug step hardware of the system (or simulating the actions of that hardware) then the instruction are executed, but not in sequence as they have all the steps of the debugger/tracing around them.
>

 That is not how x86 emulators work.

But it is what "Debug Step" implies.
Or, are you not familiar with that part of the x86 hardware.

 

So, your claim of what happens just shows you don't understand what the program you are using actually is doing.
>

 No it shows that you don't know how x86 emulators work.

So, what did I not understand?

 

That might explain why the trace you posted the other day wasn't actually the trace you claimed it was.
>

 We are only focusing on this one thread and zero deflection
will be tolerated.

Ok, so why doesn't your HH here trace into HH
And why does it start to show a trace of instructions that are never actually directly simulated by the HH that we are talking about>

 

>
This is the only post that I will reply to you on.
I need you to stay focused on this one single point
until you understand it.

>
Is that a promise? I think you will break it.
>

 When you proved to break out of your "stuck in rebuttal mode"
nonsense and talked about closure I backed off this requirement
for a while.
 *Even Ben admits that H does meet the Sipser criteria*
 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.
 

Nope, he is just pointing out your circular logic.
Yes, P(P) won't halt if the code of H won't abort it.
But if the code of H does abort it, it will halt.
Thus, if the question actually WAS "Will the template P(P) Halt if the decider it is pair with doesn't abort its simulation", then he agreed you could answer No.
But that isn't the Halting Question, so it isn't the correct answer to the question that H is supposed to answer.