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computers / comp.ai.philosophy / Re: Is it possible to create a simple infinite emulation detector? [ cite sources ]

SubjectAuthor
* Is it possible to create a simple infinite emulation detector?olcott
`- Re: Is it possible to create a simple infinite emulation detector? [olcott

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Subject: Is it possible to create a simple infinite emulation detector?
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, comp.lang.c, comp.lang.c++
Date: Thu, 23 Sep 2021 18:30 UTC
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From: NoO...@NoWhere.com (olcott)
Subject: Is it possible to create a simple infinite emulation detector?
Date: Thu, 23 Sep 2021 13:30:14 -0500
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#include <stdint.h>
#define ptr uintptr_t

int H(ptr p, ptr i)
{
// Determine infinitely nested x86 emulation
}

void P(ptr x)
{
   H(x, x);
}

int main()
{
   printf("H is called in infinitely nested emulation = ", H(P, P));
}

H would use an x86 emulator to emulate its input in debug step mode.

Since people are telling me that my solution is incorrect I am giving them an opportunity to either correct my errors or failing that show that their software engineering skills are insufficient to analyze the problem as presented.


--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre minds." Einstein


Subject: Re: Is it possible to create a simple infinite emulation detector? [ cite sources ]
From: olcott
Newsgroups: comp.theory, comp.ai.philosophy, sci.logic, sci.math
Date: Sun, 26 Sep 2021 03:00 UTC
References: 1 2 3 4 5 6 7 8 9 10 11 12 13 14
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NNTP-Posting-Date: Sat, 25 Sep 2021 22:00:36 -0500
Subject: Re: Is it possible to create a simple infinite emulation detector? [
cite sources ]
Newsgroups: comp.theory,comp.ai.philosophy,sci.logic,sci.math
References: <eP2dnfJCwLgqWtH8nZ2dnUU7-YPNnZ2d@giganews.com>
<dra3J.78902$Dr.49576@fx40.iad>
<2NydnbC8WNAD2ND8nZ2dnUU7-f3NnZ2d@giganews.com>
<eMb3J.17462$YG4.1741@fx15.iad>
<no2dnRC5q77teND8nZ2dnUU7-KnNnZ2d@giganews.com> <87lf3ly9r2.fsf@bsb.me.uk>
<PJ-dndR6J8ZQ1dP8nZ2dnUU7-cNQAAAA@giganews.com> <87v92pwod0.fsf@bsb.me.uk>
<g46dnfOiA4B6-9P8nZ2dnUU7-UXNnZ2d@giganews.com> <87r1dcv12c.fsf@bsb.me.uk>
<Pb-dneflRqJ6F9L8nZ2dnUU7-audnZ2d@giganews.com> <87fstsuz5q.fsf@bsb.me.uk>
<OdOdnX3XffppOtL8nZ2dnUU7-cvNnZ2d@giganews.com> <87tui8tdl8.fsf@bsb.me.uk>
From: NoO...@NoWhere.com (olcott)
Date: Sat, 25 Sep 2021 22:00:34 -0500
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On 9/25/2021 6:55 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 9/25/2021 4:24 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 9/25/2021 3:43 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 9/24/2021 6:22 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

On 9/24/2021 3:54 PM, Ben Bacarisse wrote:
olcott <NoOne@NoWhere.com> writes:

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt
This is wrong since, apparently,
       "I always used this symbol Ĥ in the context of the Linz proofs and
       never used it in any other way"
If Ĥ refers to the Ĥ in Linz's proof your annotations are incorrect, and
if Ĥ is being using "in the context of the Linz proof" but denotes
something else (what?) then you are not being serious.

As you have indicated that you have recalled I corrected the error in
the Linz specification that had two start states by changing one of
these start states to qx.
Not the issue.  (Though I'll note again that you don't use the notation
you came up with in a helpful way.  There is no need for qx because it
could be called H.q0 or, better yet, H.0.)

After this I augmented the Linz notation further so that it showed the
notation when the Linz Ĥ is specifically applied to the Linz ⟨Ĥ⟩ where
the halt decider embedded at qx is a simulating halt decider.
Not the issue but if you used the . notation better this would be much clearer.

Since Linz does not exclude simulating halt deciders this is still the
Linz Ĥ.

You annotations are incorrect.  If you'd like to know why, ask an
intelligent question.  But why are you making any claims at all about
TMs that don't exist?

How would you annotate all of the steps of the Linz Ĥ applied to its
own machine description?
What new nonsense is this?  No one is annotating "all of the steps" of a
TM.  The annotations explain which formal statements apply in which
situations:
     Ĥ.q0 ⟨Ĥ⟩ ⊢* ∞      if Ĥ applied to ⟨Ĥ⟩ halts, and
     Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qn   if Ĥ applied to ⟨Ĥ⟩ does not halt.

You skipped some Linz specified states.
Your annotations are still wrong.  Mine (in fact Linz's) are the correct
ones regardless of whether I (or he) includes the states you are so
obsessed with.

As always you can dogmatically assert that I am wrong yet cannot
provide any evidence of my error because there is in fact no error.

What would be the point in copying here what Linz says in his book
again?  I have, in fact, gone through the steps from the definition of
the problem to the two lines above but it had no effect.  It's right
there in the book -- the simple steps from definition

   H.q0 w ⊢* H.qy   if H applied to w halts, and
   H.q0 w ⊢* H.qn   if H applied to w does not halt,

through the construction of H' and H^ to the conclusion above.  Do you
think you'd pay attention this time if I went though it all again?  No,
you just want to pretend that there is no "evidence" for what I say.

Have another go.  See if you can write the two lines about Ĥ.q0 ⟨Ĥ⟩
correctly with intermediate states you love so much.

When the halt decider at Ĥ.qx is a simulating halt decider

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qy ∞
if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ halts, and

Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.qx ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn
if the simulated ⟨Ĥ⟩ applied to ⟨Ĥ⟩ does not halt

Provided the first clause applies "if Ĥ applied to ⟨Ĥ⟩ halts", and the
second one "if Ĥ applied to ⟨Ĥ⟩ does not halt", you can add any old guff
you like.  So just make sure you say something like


That is not what it says and that is not what Linz says.

The halt decider is only at state Ĥ.qx and is only applied to two copies of the TM description of Ĥ it cannot be applied to Ĥ itself, the input must be a TM description and not a TM.

When the halt decider at Ĥ is a simulating halt decider then it transitions to Ĥ.qn on the basis that the simulation of its first TM description ⟨Ĥ⟩ applied to its second TM description ⟨Ĥ⟩ never reaches its final state whether or not the simulating halt decider at Ĥ.qx ever stops simulating this input.



--
Copyright 2021 Pete Olcott

"Great spirits have always encountered violent opposition from mediocre minds." Einstein


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