From: Peter de Blanc (firstname.lastname@example.org)
Date: Mon Jun 23 2008 - 00:42:24 MDT
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
> finite machine. Either way, you have a problem.
Turing machines have infinite memory. But the argument still works for
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
> 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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