Oo, this is a good one. Let's do some analysis:
We can see the centres of three sides, and the relative positions of the centres cannot be changed, so we know that when/if this cube is solved, the blue side will be adjacent to green and orange.
We can also see a blue-white edge piece, and a blue-yellow edge piece as well. This means that as long as the cube ...
The question is a bit unclear - it sounds like you're asking "Given a sequence of moves A, is there always some number of repetitions of A that will get you back to the start state?"
This is relatively easy to prove - it's just basic group theory.
Imagine you have a bunch of Rubik's cubes -- one in every possible state. Given an algorithm A, you ...
Okay, I'll give a group-theoretic answer, since that was requested, which goes most of the way to a proof. I'm assuming you mean a 3x3x3 Rubik's cube. For larger cubes, the proof is just a little bit less convenient, because these cubes don't form groups (or more precisely, the action of the group of all moves on the cube on the states of the cube is not ...
There are three techniques that allow you to come up with useful move sequences for solving a cube.
This is where you already have a move sequence that does one thing and allows you to apply it more generally.
Suppose for example that you know a move sequence X that twists the two adjacent corners at the front of the top layer (i.e. at UFR and ...
The answer is
I do hope the explanation and pictures suffice with no real need for a geometrical argument, because I'm not sure I can give a rigorous proof.
This solution was found by Robert S. Holmes as part of a challenge called "a set of quickies" by Martin Gardner (who originally had found a solution of 20). The problem and Martin Gardner's 20-...
You shouldn't think about the cube in terms of tiles, but on the mini-cubes it is split into, sometimes called cubies or cubelets. There are 26 of these (3*3*3 minus the center), and there are 3 types. The centers (1 color), the edges (2 colors), and the corners (3 colors). For instance, it is not just a blue tile, it is a blue-red edge piece, which there is ...
If you're willing to take the time: The World Cube Association has a specific method for shuffling their cubes. They use a program called TNoodle to do the scrambling; the program generates a "scramble sequence" which can be followed to produce a scrambled cube.
Their code actually features a nice interface to generate the scrambles:
I have yet to figure ...
My answer at Why is a single-corner twist not a valid position on a Rubik's cube? basically explains this, but in simpler terminology here are the facts presented without proof.
If you randomly disassemble and reassemble a $3 \times 3 \times 3$ cube, the probability that it will be solvable with just face turns is exactly $1 \over 12$. Furthermore, this ...
In a Rubik's cube, every legal move swaps a even number of dowels, so any legal configuration can be obtained only with a even number of swaps.
In this configuration, the difference between a legal cube (the solved one) and the current status consists of 1 swap; since 1 is odd, this is a No Win Scenario.
The "most scrambled cube" is any configuration that requires the greatest number of rotations to solve the cube using a perfect algorithm.
Although unknown, this algorithm is hypothetically called God's algorithm and in fact the maximum number of rotations, called God's number has been found to be 20.
The list of such 20-move cubes is given here.
A Rubik's cube has six faces, so there are six different colors to work with.
This suggests to me that it's
So I changed each of the dates into
Thus the colors represent:
Read the pattern from the 3rd block:
So my answer is:
I'm not sure the significance of this day, if there is one.
Each set of moves is designed to correct a parity error in the cube's layout, but each group of parity it corrects is essentially independent and random. Each step has a chance of already having the right parity. For instance, the way I do it is solve the bottom two layers and then attack the parity errors on the top one at a time
Make sure all the edges ...
It seems like you are trying to find
One such example that satisfies this is
As mentioned by armb in the comments, there is a good answer here discussing the maximum orders for an $n \times n \times n$ Rubiks cube.
The World Cubing Association actually has an entire event dedicated to solving the cube with the fewest move count (FMC). You are given a scramble, paper, pencil, up to 3 cubes and an hour.
The methods used vary widely and involve much more "freestyle" than regular speedsolving. But I'll explain one of the more common "methods." (Which is more like a set of ...
The easiest explanation would be that in a 3x3 cube, only one cube is out of position, but in a 4x4 cube two cubes are out of position.
In a 15 puzzle (the sliding puzzle where you try to put the numbers in order) half of all possible initial positions are unsolvable. They call the solvable positions "even" and the unsolvable positions "odd". The "odd" ...
They simply decide on 98 different valid Rubik's cube states at the beginning of the game and denote them 1 to 98. On the first day, the prisoner leaves the cube in any state apart from the pre-defined 98. We can call this state 0.
After this, everyone (including the 1st prisoner) follows the same strategy. Whenever a prisoner sees the cube for the first ...