From: J. Andrew Rogers (andrew@ceruleansystems.com)
Date: Sun Jun 22 2008 - 20:04:02 MDT
On Jun 22, 2008, at 6:23 PM, Peter de Blanc wrote:
> Suppose you have a machine that outputs a copy of itself, except that
> somewhere in its memory it contains a model number, and each time it
> makes a copy, it increments the model number on the copy.
>
> This machine and all of its descendants comprise an infinite set of
> distinct machines. Call the set S. Let K be the complexity of
> machine 1.
> Since there are only finitely many machines of complexity K or less, S
> must contain some machine with complexity greater than K. So let N be
> the smallest number such that machine N has complexity greater than K.
> Then machine N-1 has complexity less than or equal to K, so machine
> N-1
> is a machine that outputs a machine of greater algorithmic complexity
> than itself.
Eh, you might want to double-check your reasoning; it could be picked
apart at a couple points where you make an inconsistent or unwarranted
leap. It appears that, at a minimum, you failed to grasp how the
counter function interacts with the complexity of the machine.
J. Andrew Rogers
This archive was generated by hypermail 2.1.5 : Wed Jul 17 2013 - 04:01:03 MDT