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

**Date:** Mon Jun 23 2008 - 11:38:01 MDT

**Next message:**Peter de Blanc: "Re: [sl4] Is there a model for RSI?"**Previous message:**J. Andrew Rogers: "Re: [sl4] Is there a model for RSI?"**In reply to:**William Pearson: "Re: [sl4] Is there a model for RSI?"**Next in thread:**Peter de Blanc: "Re: [sl4] Is there a model for RSI?"**Reply:**Peter de Blanc: "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 ]

William Pearson wrote:

*> Which proof are you talking about here?
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The one in my first post in my thread, at 6/22/2008 6:23 PM pacific time.

*> I am trying to tell you that
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*>
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*> "Since there are only finitely many machines of complexity K or less"
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*>
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*> Is incorrect, if you use chaitin or kolmogorov complexity.
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*>
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*> Will Pearson
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According to Wikipedia's definition of Kolmogorov complexity, the

statement is true.

But I was saying that if you substitute "finitely-many algorithms" for

"finitely-many machines", then the proof still works given your definition.

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