**From:** Vladimir Nesov (*robotact@gmail.com*)

**Date:** Mon Jun 23 2008 - 05:03:55 MDT

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On Mon, Jun 23, 2008 at 9:38 AM, J. Andrew Rogers

<andrew@ceruleansystems.com> wrote:

*> If your "counter" does not occupy memory then the complexity is constant.
*

*> If it does occupy memory, it is eating bits as N grows on a finite machine.
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*> Either way, you have a problem.
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*>
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*> All that aside, your assertion is contrary to some pretty rudimentary (for
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*> this list) theorems in mathematics. Without inspecting your argument, that
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*> alone should have suggested something was amiss.
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*>
*

No, Peter is correct, and it is possible to construct even simpler

example. Say, our universal TM that runs the show only stops for two

programs of length less than 10: for P1 of length Len(P1)=3 and for P2

with Len(P2)=4. You can encode any other TM to run on this UTM, but

it'll be at least of length 10: for all P that stop, P<>P1, P<>P2,

Len(P)>=10. Let's say, the output of running P1 is Run(P1)=P3, and

output of P2 is Run(P2)=P4, but we make sure that also Run(P3)=P4.

Then, clearly Kolmogorov complexity of P3 is 3 (length of P1), and of

P4 it's 4 (length of P2), but P3 outputs P4, and P4 has greater

Kolmogorov complexity than P3.

-- Vladimir Nesov robotact@gmail.com http://causalityrelay.wordpress.com/

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