AlephNeil comments on A simple counterexample to deBlanc 2007? - Less Wrong

3 Post author: PhilGoetz 30 May 2011 05:09AM

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Comment author: AlephNeil 30 May 2011 08:04:46AM *  0 points [-]

Because then you can just number your possible outcomes by integers n and set p(n) to 1/U(n) * 1/2^n, which seems too easy to have been missed.

The reason why this wouldn't work is that sometimes what you're calling "U(n)" would fail to be well defined (because some computation doesn't halt) whereas p(n) must always return something.

Comment author: PhilGoetz 30 May 2011 03:05:25PM *  0 points [-]

The reason why this wouldn't work is that sometimes what you're calling "U(n)" would fail to be well defined (because some computation doesn't halt)

No; the utility function is stipulated to be computable.

Comment author: AlephNeil 30 May 2011 03:09:52PM *  0 points [-]

No; the utility function is stipulated to be computable.

What Manfred is calling U(n) here corresponds to what the paper would call U(phi_n(k)).

Comment author: PhilGoetz 30 May 2011 06:12:14PM 0 points [-]

The utility function is defined as being computable over all possible input.

Comment author: AlephNeil 31 May 2011 11:44:54AM 0 points [-]

phi_n(k) may not halt.

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