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Khoth comments on Open Thread March 31 - April 7 2014 - Less Wrong Discussion

2 Post author: beoShaffer 01 April 2014 01:41AM

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Comment author: RichardKennaway 01 April 2014 07:28:20PM 1 point [-]

Doesn't Cantor's diagonal argument prevent you from having a countable ordering of all the sequences?

There's no need for an enumeration of all the sequences, bayl na rahzrengvba bs gur rdhvinyrapr pynff gung gur cevfbaref frr gung gurl'er va jura gur ungf ner cynprq. Naq Bfpne_Phaavatunz'f fbyhgvba qbrfa'g arrq rira gung -- gur rdhvinyrapr pynffrf pna or nal genafsvavgr fvmr jungrire.

I'm now wondering whether for the case of two colours, there is a computable algorithm. A prisoner would apply the algorithm by feeding it an infinite tape listing all the colours of the hats, with a blank for his own, and the algorithm would in a finite time say what guess to make.

Comment author: Khoth 01 April 2014 10:18:32PM 0 points [-]

It seems unlikely. In a finite time it's impossible to get any idea whatsoever what equivalence class they're in, so the solution, if there is one, would need to be very different.