RobinZ comments on Frequentist Magic vs. Bayesian Magic - Less Wrong

41 Post author: Wei_Dai 08 April 2010 08:34PM

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Comment author: gwern 13 April 2010 01:16:30AM 1 point [-]
  1. Finding BB(n) for all n is uncomputable. Finding BB(n) where n=10, say, is not uncomputable.
  2. Yeah, this is an interesting point. My thinking is that yes, we do only care about modeling Turing machines that fit in the universe, since by definition those are the only Turing machines whose output we will ever observe. We might also have to bite the bullet and say we only care about predicting Turing machines which are small enough that there is room left in the universe for us; we don't want to perfectly model the entire universe, just small parts.
Comment author: RobinZ 15 April 2010 08:28:46PM *  1 point [-]

Just out of my own curiosity, I checked Wikipedia on Busy Beaver numbers: BB(10) > 3^^^3.

While the article admits that every finite sequence (such as n = 1...10) is computable, I think it is safe to call it intractable right now.

Comment author: gwern 15 April 2010 10:11:10PM 2 points [-]

Hey, don't laugh. That's useful to know - now you know any Turing Machine less than 10 is bluffing when it threatens 3^^^^3 specks of sand in eyes.

Comment author: JGWeissman 15 April 2010 11:46:24PM 1 point [-]

now you know any Turing Machine less than 10 is bluffing when it threatens 3^^^^3 specks of sand in eyes.

How did you conclude that from a lower bound?

Comment author: gwern 16 April 2010 02:17:44AM 2 points [-]

I don't think the lower-bound will turn out to be all that off, and I tossed in another ^ as insurance. Jests don't have to be air-tight.

Comment author: thomblake 15 April 2010 10:30:52PM 1 point [-]

Off the cuff, couldn't it actually accomplish that if it just doesn't terminate?

Comment author: gwern 15 April 2010 11:38:28PM 1 point [-]

If it doesn't terminate, then I think its threat would be an infinite number of sand specks, wouldn't it?