[next] [prev] [up] Date: Fri, 20 Oct 95 21:27:27 +0000 (GMT)
[next] [prev] [up] From: Wei-Hwa Huang <whuang@cco.caltech.edu >
~~~ [prev] [up] Subject: Re: Old question about 2 adj edges

Mikko Haapanen <hazard@niksula.hut.fi> writes:
>This reminds me another old question: 3x3x3 are told to have about 4
>trillion (or whatever) different positions. How many of these positions are
>'solved cube' but with different centerpiece combinations? Once i had 3x3x3
>with 6 different pictures (picture/side). Friends asked me to solve it. When
>i was completed, they laughed at me and pointed the bottom center piece,
>which was out of orientation (i can't remember how many of centers were out
>of order).

Actually, I think the 4 "trillion" estimate is ignoring the center orientation.
Let's see:
8! corner positions
x 3^7 corner orientations
x 12!/2 edge positions
x 2^11 edge orientations
= 4.325x10^19
Well, forty-three quadrillion.

Five center orientations force the sixth, so multiply your number by 4^5 to
get the answer 4.429x10^22 positions, counting center piece
orientations. That's 44 quintillion. Whew.

I remember when I solved the 5x5x5 cube (finally), someone asked me if I
had solved the "invisible" 3x3x3 inside it. I'm not sure I even want to
think of trying to solve that. I'll work on the 3x3x3x3 first. :P

int m,u,e=0;float l,_,I;main() {for(;e<1863; putchar((++e>924&&952>
e?60-m:u) ["\n  ude.hcetlac.occ@gnauhw   ]"])) for(u=_=l=0;(m=e%81)
<80&&I*l+_*_ <6&&25>++u;_=2*l*_+e/81*.09-1,l=I)I=l*l-_*_-2+.035*m;}

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