I was feeling mathematical today and was curious if I am thinking correctly (my math skills aren't very advanced due to only being in Algebra 2).
Here is an equation I made made from memory and logic so I wouldn't be surprised if it isn't correct.
12! x 2^10 x 8! x 3^7
I assume that this is correct since it yields 4.325x10^20 in my calculator
I was using this equation to try and find the odds of have a solvable cube when disassembled and reassembled randomly.
I got 3^7 from realizing that you can't rotate one of the corners without affecting another (ie 8 corners minus one that is unmovable=7) and the fact that you can't swap two edges without affecting the corners (ie 12 edges-2=10)
Since these rules don't apply in random assembly I came up with 3^8 and 2^12.
2^12-2^10 equals 2^2 or 4.
3^8-3^7=3^1 or 3.
4 times 3 equals 12. If I take that twelve and multiply it by the original number of scrambles I get 12 times as many scrambles (duh!). So that means that only one in twelve assembled scrambles will also be a solvable scramble.
I have a Type A-I, and I'm wondering if the A-II is much different from it. Should I just get the other five, or is the A-II good enough that it's worth getting in addition to the others?
I have a Type A-I, and I'm wondering if the A-II is much different from it. Should I just get the other five, or is the A-II good enough that it's worth getting in addition to the others?
I have a Type A-I, and I'm wondering if the A-II is much different from it. Should I just get the other five, or is the A-II good enough that it's worth getting in addition to the others?
I love my Type a II. Doesnt pop, fast, cuts corners. The only problem is the lockups. Imma let you finish, but Ghost Hands are some the best cubes of ALL Time.
I was feeling mathematical today and was curious if I am thinking correctly (my math skills aren't very advanced due to only being in Algebra 2).
Here is an equation I made made from memory and logic so I wouldn't be surprised if it isn't correct.
12! x 2^10 x 8! x 3^7
I assume that this is correct since it yields 4.325x10^20 in my calculator
I was using this equation to try and find the odds of have a solvable cube when disassembled and reassembled randomly.
I got 3^7 from realizing that you can't rotate one of the corners without affecting another (ie 8 corners minus one that is unmovable=7) and the fact that you can't swap two edges without affecting the corners (ie 12 edges-2=10)
Since these rules don't apply in random assembly I came up with 3^8 and 2^12.
2^12-2^10 equals 2^2 or 4.
3^8-3^7=3^1 or 3.
4 times 3 equals 12. If I take that twelve and multiply it by the original number of scrambles I get 12 times as many scrambles (duh!). So that means that only one in twelve assembled scrambles will also be a solvable scramble.
I ordered from popbuying on 2010/01/02, it "shipped" on 2010/01/13 but left the post office on 2010/01/19 what do you think when will i get it?
I'm located in Germany btw.
(no i'm not impatient, just curious)
I ordered from popbuying on 2010/01/02, it "shipped" on 2010/01/13 but left the post office on 2010/01/19 what do you think when will i get it?
I'm located in Germany btw.
(no i'm not impatient, just curious)
In keyhole, you insert corners first then edges. Does anyone insert edges first then corners?
Why don't more people do cross on left. In solves the only faces I have to use are l, R, U, & x. With cross on bottom I use R, U, L, F, y, & M. Isn't cross on left more finger-tricky?