**From:** Peter de Blanc (*peter@spaceandgames.com*)

**Date:** Mon Jun 23 2008 - 00:42:24 MDT

**Next message:**J. Andrew Rogers: "Re: [sl4] Is there a model for RSI?"**Previous message:**J. Andrew Rogers: "Re: [sl4] Is there a model for RSI?"**In reply to:**J. Andrew Rogers: "Re: [sl4] Is there a model for RSI?"**Next in thread:**J. Andrew Rogers: "Re: [sl4] Is there a model for RSI?"**Reply:**J. Andrew Rogers: "Re: [sl4] Is there a model for RSI?"**Reply:**William Pearson: "Re: [sl4] Is there a model for RSI?"**Messages sorted by:**[ date ] [ thread ] [ subject ] [ author ] [ attachment ]

J. Andrew Rogers 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
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*> finite machine. Either way, you have a problem.
*

Turing machines have infinite memory. But the argument still works for

finite machines:

Let K be the complexity of machine 0. Let M be the number of machines of

complexity at most K (if complexity is measured in bits, that's

2^(k+1)-2). Then somewhere in the set 0...M, there's a machine with

complexity greater than K. We're only counting up to M, so that's pretty

finite.

*> All that aside, your assertion is contrary to some pretty rudimentary
*

*> (for this list) theorems in mathematics.
*

So show us these alleged theorems.

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