Re: Why does Olcott care about simulation, anyway? --- Mikes Review

On 6/3/24 10:18 PM, olcott wrote:

On 6/3/2024 8:59 PM, Richard Damon wrote:

On 6/3/24 9:46 PM, olcott wrote:

On 6/3/2024 8:38 PM, Mike Terry wrote:

On 03/06/2024 18:54, olcott wrote:

On 6/3/2024 11:25 AM, Mike Terry wrote:

On 03/06/2024 04:50, olcott wrote:

On 6/2/2024 10:28 PM, Mike Terry wrote:

On 03/06/2024 01:16, immibis wrote:

The halting problem says you can't find a Turing machine that tells whether executing each other Turing machine will halt. Simulation has nothing to do with the question.

>
Background:
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PO claims to have refuted the common HP proof, e.g. as covered in the Linz book "An Introduction to Formal Languages and Automata". PO occasionally posts a link to a PDF containing an extract of the 5 or so pages of the book containing the proof, but I expect you have access to this or equivalent.
>
In a nutshell, the proof goes:
-  Suppose H is a TM Halt decider that decides for any input <P><I> whether
    TM P run with input I on its input tape halts.
    [<P> is the string representation of the actual TM P, and
     <I> is the string representation of input tape I]
-  Construct from H a new TM H^ using the mechanical process described in the proof.
    If H exists, then its corresponding H^ also exists.
-  Show that the construction of H^ ensures that:
    -  if H decides input <H^><H^> (representing H^ running with input <H^>) halts,
       then that implies that H^ running with input <H^> never halts
    -  if H decides input <H^><H^> never halts,
       then that implies H^ running with input <H^> halts
    I.e. either way, H decides the specific input <H^><H^> incorrectly, contradicting
    the initial assumption that H is a halt decider.
-  So no halt decider exists.  (Every proposed halt decider decides at least one input case
    incorrectly, viz input <H^><H^>.)
>
PO basically claimed he had a fully coded TM H, which CORRECTLY decides its "nemesis" input <H^><H^>, contradicting the logic of the Linz proof [without pointing out any actual mistake in the Linz proof].  Given most people here understand the Linz proof well enough to see it is basically sound, people were sceptical!
>
It turned out PO was lying about the fully coded TM, and in fact what he actually had was the idea behind a C program which would "prove" his idea.  A couple of years(?) later he actually completed his C program and his x86utm.exe which would simulate the x86 code of his H and H^ to "prove" his claim.  His equivalent of Linz H is his C function H or HH, and his equivalent of Linz H^ is his D or DD respectively.  (They run under x86utm.exe and are not Windows/Unix executables.)
>
H/HH use PARTIAL simulation of their input to decide halting/non-halting, returning
0 or 1 to communicate their decision.  As you correctly point out, to the HP proof simulation is quite irrelevant, being just one kind of data manipulation that H may perform on its input string <P><I> before it decides the halting status.  So the Linz HP proof covers such H, no problem.
[I put PARTIAL in caps, just because there seems to be some confusion in recent threads as to what PO means by "simulation". He doesn't say it explicitly, despite suggestions to this effect, but he always means what might be called /partial/ simulation.]
>
PO believes that by (partially) simulating the computation corresponding to the input <P><I> [i.e. calculating the successive x86 instruction steps of the computation P(I)] and monitoring the progress of virtual x86 state changes (like instruction address and op-code and so on) H could spot some pattern that reveals whether computation P(I) halts or not.  At this point in the partial simulation, his H would stop simulating (aka "abort" the simulation) and return the appropriate halt status for input <P><I>.
>
Nothing remarkable so far!  Clearly a tight-loop in P /can/ be detected in this fashion, so /some/ <P><I> inputs /can/ be correctly determined like this.  The PO claim however is that the specific input <H^><H^> is correctly decided by his H.  In C terms those correspond to H(D,D) correctly returning the halt status of computation D(D).  [PO would probably dispute this, because he doesn't properly understand halting or the HP generally, or in fact pretty much /any abstract concept/ ]
>

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Introduction to the Theory of Computation, by Michael Sipser
https://www.amazon.com/Introduction-Theory-Computation-Michael-Sipser/dp/113318779X/
>
On 10/13/2022 11:29:23 AM
MIT Professor Michael Sipser agreed this verbatim paragraph is correct
(He has neither reviewed nor agreed to anything else in this paper)
>
<Professor Sipser agreed>
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.
</Professor Sipser agreed>
>
I have started working on what seem to be some computability issues
that you pointed out with my HH. I found that they are isolated to
one single element of HH. Essentially the details of how the master
UTM directly executed HH passes a portion of its tape to its slaves.
>
Nothing else seems to have any computability issues by the measure
that I am using.
>
Message-ID: <rLmcnQQ3-N_tvH_4nZ2dnZfqnPGdnZ2d@brightview.co.uk>
On 3/1/2024 12:41 PM, Mike Terry wrote:
 >
 > Obviously a simulator has access to the internal state
 > (tape contents etc.) of the simulated machine. No problem there.
>
Because of your above comment it seems that correcting this
tiny computability issue with HH is my best path forward.
>
>
>
>
You have given the following a blatantly false review when I
said the same thing another way:

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I have no idea what you're talking about.  I did not write any of what follows below.
>
Also I don't believe I said anything "blatantly false".  You're incapable of judging what other people say or are thinking - you're often telling people that they'er lying to you and denying
"previously verified facts" etc. but its all rubbish - you're in no position to make such judgements.
>
>
Mike.
>

>
You said that the execution trace that I proved is correct is
incorrect because you didn't like the way that HH was written.
You said this without looking at my proof as you are doing
here again.

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An execution trace that is produced by a program that is incorrect /proves/ nothing whatsoever.  I don't need to look at your proof, as I was commenting on the value of your program output AS PROOF.
>

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I provided the execution trace that HH derives
*AND THE X86 SOURCE-CODE OF DD THAT PROVES THIS TRACE IS CORRECT*
*AND THE X86 SOURCE-CODE OF DD THAT PROVES THIS TRACE IS CORRECT*
*AND THE X86 SOURCE-CODE OF DD THAT PROVES THIS TRACE IS CORRECT*

>
Then why did the trace not follow the call to H?
>

 HH(DD,DD) the trace does follow the call to HH(DD,DD)
and fully simulates itself simulating DD.

So, where are the instuctions of HH shown?
I guess you are just a LIAR.

 DD correctly emulated by any HH that can possibly exist DOES NOT HALT
DD correctly emulated by any HH that can possibly exist DOES NOT HALT
DD correctly emulated by any HH that can possibly exist DOES NOT HALT

PROVEN WRONG, and that you are just a STUPID IDIOT THAT DOESN'T KNOW THE MEANING OF THE WORDS.
DD DOES HALT, as Halting is the behavior of the machine itself.
If you keep repeating your lie, I will just keep focusing on the error of you not knowing the meaning of the word.

 typedef int (*ptr)();  // ptr is pointer to int function in C
00       int HH(ptr p, ptr i);
01       int DD(ptr p)
02       {
03         int Halt_Status = HH(p, p);
04         if (Halt_Status)
05           HERE: goto HERE;
06         return Halt_Status;
07       }
 _DD()
[00001c22] 55         push ebp
[00001c23] 8bec       mov ebp,esp
[00001c25] 51         push ecx
[00001c26] 8b4508     mov eax,[ebp+08]
[00001c29] 50         push eax        ; push DD 1c22
[00001c2a] 8b4d08     mov ecx,[ebp+08]
[00001c2d] 51         push ecx        ; push DD 1c22
[00001c2e] e80ff7ffff call 00001342   ; call HH
[00001c33] 83c408     add esp,+08
[00001c36] 8945fc     mov [ebp-04],eax
[00001c39] 837dfc00   cmp dword [ebp-04],+00
[00001c3d] 7402       jz 00001c41
[00001c3f] ebfe       jmp 00001c3f
[00001c41] 8b45fc     mov eax,[ebp-04]
[00001c44] 8be5       mov esp,ebp
[00001c46] 5d         pop ebp
[00001c47] c3         ret
Size in bytes:(0038) [00001c47]