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Rubik's cube combos

T

TK

Member
Standard cube - pamphlet says about 43 quintillion, seemed like a lot! I attempted to calculate and came up with even more - 519 quintillion. Seems like a fairly simple (4 part) equation: fact. 8 (for corner locations) X fact. 12 (for edge locations) X 3^8 (for corner colors) X 2^12 (for edge colors). What am I missing (or adding)?
 
Not all combinations are attainable.

Example: if you take a cube apart and put it back together wrong (ie 1 corner piece is twisted wrong, but all other pieces are in their correct positions) then the cube is unsolvable.
 
smack down - as each piece is unique, I don't see how there can be any similar positions.

confused - I have had 2 or 3 cubes over the years, and each came with a small booklet that tells how to flip and position all the pieces for the solution (brute force).

Jaepheth - I would like to disagree with you, but I believe you are probably correct! That gives me more to consider, and it would certainly reduce the number I've been getting. I'll play with it some.
 
I think what smack down means is that turning the whole cube without changing any relative positions would technically be a different combination, but we would consider them the same.
 
Originally posted by: Jaepheth
I think what smack down means is that turning the whole cube without changing any relative positions would technically be a different combination, but we would consider them the same.
Click to expand...

That is accounted for because he didn't include anything about center piece locations in the equation -- what matters is the locations of other pieces relative to the centers of the sides. Note that it is impossible to change the arrangement of the centers of the sides relative to each other.


The impossible edge and corner rotations you are missing are explained on Wikipedia
 
glugglug - That answers the number question (equation). Now I want to see "feel?" which combinations won't work. Why (3^8-1 and 2^12-1)/2,

Good link, thanks.
 
Yea in a 3x3x3 cube having the centers fixed relative to one another is really useful (and although i never really thought about it permutation limiting). I just moved to a 4x4x4 cube, which has non-fixed centers and for a while i was really lost. In fact in order for me to solve the 4x4x4 I HAVE to solve the centers first, i cant get my head around it any other way.

maybe this is a little off topic, but i've been having a problem with solving my 4x4x4 that maybe someone can help me with. I solve for the centers, match up all the middles (l's and r's) , and solve like a regular cube. I come to a situation where i have two opposing faces solved and one non-opposing face solved. That leaves only two sets of middles out of place (but still matched). I am familiar with this situation from the 3x3x3 and can solve that cube using this certain combination i learned (its one i havent seen online and have trouble describing using typical notation: the cube is held so that the two swaped middles are in the U face, then i rotate the center slice clockwise (the slice between R and L in standard notation), U, center slice, U, center slice, U", center slice counter clockwise, U, center slice counter clockwise, U, center slice counter clockwise, U"). However when i apply the same combination to my 4x4x4 it only rarely works, but it never messes the cube up either. I'm looking for someone who can explain what is happening or who can offer a substitute sequence to try.
 
Originally posted by: TK
glugglug - That answers the number question (equation). Now I want to see "feel?" which combinations won't work. Why (3^8-1 and 2^12-1)/2,

Good link, thanks.
Click to expand...

The 3^8-1 is because you can't turn just one corner -- they have to be rotated in pairs. So anything that is only 1 turned corner from a possible combination is not possible, and it means if you know the positions and angles of 7 corners, the 8th corner this limits the last corner to only 1 angle.

I also get the 2^12-1 is because you can't flip just one edge -- the edges also have to be flipped in pairs.

The final /2 is because since you can't swap only 2 edge pieces, so if the locations of 10 edge pieces are known, this gives only 1 possibility for the remaining 2 edge pieces.
 
Originally posted by: Jaepheth
Not all combinations are attainable.

Example: if you take a cube apart and put it back together wrong (ie 1 corner piece is twisted wrong, but all other pieces are in their correct positions) then the cube is unsolvable.
Click to expand...

actually, i have solved it several times with a corner piece deliberately installed incorrectly. maybe i just got lucky, but ive done it several times.
 
You definately just got lucky. Solve the cube all the way, switch a single corner and try and solve it. I've done it and found the cube to be impossible to solve.
 
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