Good morning all..
Itsalready!!![]()
Wake up and get it over with!!
A few things while I'm here. First the good news, then the bad news.
Good News:
-I got my PS3 today.![]()
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It doesn't work on my TV, but worked on another, so I need to get a new TV! Nothing fancy, just one with the r/y/w ports in front and a working remote. My current TV is an old one, so it's upgrade time!![]()
-The parks have been slow, so anyone going to the parks will enjoy short lines.![]()
Bad News:
-Hannah and Ike are trying to give FL a scare. To them I say "Bring it, cheesewagons!"
-The DIS won't let me use "invisible text." I was going to put it as such, but you would still be able to read it. I'll just say a Split may be Imminent. PM for Details...
I have my 3 V-cubes here, and I looked at my 5cube. The 6 types of pieces make a triangle. The same thing happens with my 6cube. That made the first part really easy. Half the number of layers (ceil if it's odd so you include the middle layer) and triangulate the number.
The second part was a little more annoying. I knew there was a "mirror" piece in the 6 and 7 which didn't fit into my first equation.
The realization came when I saw that these pieces were the ones that didn't lie along the 2 major diagonals, the 2 crosses(which don't exist on evens), or the edges of a face. So I listed it out. I knew that 1-5 had to be 0. 6&7 were 1. 8&9 were 3. 10&11 were 6. (Oh look, more triangular progression). Draw it out, and you can easily see the triangle.
I came up with (order - 4)/2 ... but I liked the look of (order/2-2) better. I should probably add a lower limit of 0 for the second term.
edit:
Here's an even easier way. I'm sure there's a way to combine them.
The general idea is easy to see if you look at the 7 and go backwards to the 6.
One "Quadrant" of the 7 is a 4x4 block. Remove one leg of +centers and one leg of edges. Add the Center and Corner back in.
With the 6 you do the same thing except there are no +centers to deal with and you only add the center back in.
evens:
(order/2)^2 - (order/2) + 1
odds:
(ceil(order/2))^2 - ceil(order/2) + 1 - floor(order/2) + 1
edit 2:
(C(N/2))^2 + (1 + N%2)(1 - C(N/2)) + N%2
I still have to peel off the colors and repaste to complete a cube.
Well, if you want to know why you've never solved one if you just tried randomly....according to my son last night (again compliments of his boards!):
If every person on the earth had gotten a Rubik's cube right when they came out in 1982 and since then had made one move a second and no one made any duplications of any moves that anyone else did.....then by now still less than 1/7th of the possible outcomes would have been displayed at any time!![]()
Or, if you tried to display all the possible different ways a Rubik's cube looks, and lined themall up next to one another as close as possible, they would extend 261 light years!
The latest one he's gotten is a 7x7, which looks like this:
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But even that's too easy for him so he's just been building his own 'mutant' cubes!![]()