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On 6/2/2024 10:28 PM, Mike Terry wrote:Right, so if can show that IF H can show that CORRECTLY SIMULTED D WOULD NEVER STOP RUNNING UNLESS ABORTED.On 03/06/2024 01:16, immibis wrote:Introduction to the Theory of Computation, by Michael SipserThe 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:
>
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/ ]
>
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 issuesWhich shows part of your issue, there need not be a "master UTM" present at all, and a UTM can't effect the behavior of "its inputs" because the definition of a UTM is that it exactly recreates what the machines described by its inputs would do when they are just run.
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 measureBecause the statements are different.
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:
*We can see that the following DD cannot possibly halt when*
*correctly simulated by every HH that can possibly exist*
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]
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