Cubies constitute the different states the rubik's cube can be in. The centres of the cube never change, only the cubies are mobile and are able to change state. Upon any move in the set M, the cubies on that face perpendular to M rotate clockwise. This affects the four sides adjacent to the move by transferring the cubies on that side clockwise to the next side. for a cube to be solved, the colours of the cubies must allign with the centre (which never changes). F B R L D F and then can say stuff like RUD and that means turn the right face clockwi X means turn the face clockwise by looking direcctly at the face, and X' means turn the face counterclockwise whilelooking directly at the face R U over and over you get back solved state? what is an algorithm?: An algorithm is a finite sequence of moves the subalgorithm for a cubie, are the moves which change the state of the cubie what does it mean for an algorithm to fix a cubie? For after an algorithm, for a particular location on a cube to contain a cubie with the same orientation and colour. An algorithm fixes a cubie when the moves "applied to that cubie" cancel out.

Rubik's Cube Move Inverter

Cubie

A cubie is an atomic unit on a Rubik's Cube.

Algorithm

An algorithm is a finite sequence of moves

Rubiks Cube Base Moves

The moves of a rubiks cube

M = {f, b, r, l, u, d, e}

Move Operator

We define the move operator as follows (where the )

\forall m \in M, m^{4} = e

Also some of the elements in

M

commute

f ⋆ b = b ⋆ f

l ⋆ r = r ⋆ l

u ⋆ d = d ⋆ u

aditionally the element

e

acts like an identity:

\forall m \in M, m ⋆ e = e ⋆ m = m

States of the Cube

We define the states of the cube as a set

G

G = {\prod_{i = 1}^{n} m_{i} : m_{i} \in M and n \in ℕ_{1}}

Note that any state of the cube can be thought of as an ordered sequences of "moves" by reading from left to right which are being applied to $e$

The Rubiks Cube is a Group

The set

M

along with the move operator form a group.

Associativity:
Identity:
Let $g \in G$ . By definition of $G$ , we know there exists $m_{1}, m_{2}, \dots, m_{n} \in M$ such that $g = m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}$ for some $n \in ℕ_{1}$ . By definition of $e$ , $m ⋆ e = e ⋆ m = m$ for all $m \in M$ . So by associativity,
$\begin{matrix} e ⋆ g & a m p; = e ⋆ (m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}) \\ a m p; = (e ⋆ m_{1}) ⋆ m_{2} ⋆ \dots ⋆ m_{n} \\ a m p; = (m_{1} ⋆ e) ⋆ m_{2} ⋆ \dots ⋆ m_{n} \\ a m p; = m_{1} ⋆ (e ⋆ m_{2}) ⋆ \dots ⋆ m_{n} \\ a m p; = m_{1} ⋆ (m_{2} ⋆ e) ⋆ \dots ⋆ m_{n} \\ a m p; = m_{1} ⋆ m_{2} ⋆ \dots (e ⋆ m_{n}) \\ a m p; = m_{1} ⋆ m_{2} ⋆ \dots (m_{n} ⋆ e) \\ a m p; = (m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}) ⋆ e \\ a m p; = g ⋆ e \end{matrix}$ Therefore, $e$ is the identity element in $G$ .
Inversion:
Let $m \in M \subset G$ . Then, $m ⋆ m^{3} = m^{4} = e$ and $m^{3} ⋆ m = m^{4} = e$ . So, $m^{- 1} = m^{3}$ .

Let $g \in G$ . By definition of $G$ , we know there exists $m_{1}, m_{2}, \dots, m_{n} \in M$ such that $g = m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}$ for some $n \in ℕ_{1}$ . Now, consider $g^{'} = m_{n}^{3} ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3}$ . By associativity, we have the following. $\begin{matrix} g ⋆ g^{'} & a m p; = (m_{1} ⋆ m_{2} ⋆ \dots ⋆ m_{n}) ⋆ (m_{n}^{3} ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3}) \\ a m p; = m_{1} ⋆ m_{2} ⋆ \dots ⋆ (m_{n} ⋆ m_{n}^{3}) ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3} \\ a m p; = m_{1} ⋆ m_{2} ⋆ \dots ⋆ e ⋆ m_{n - 1}^{3} ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3} \\ a m p; = m_{1} ⋆ m_{2} ⋆ \dots ⋆ \dots ⋆ m_{2}^{3} ⋆ m_{1}^{3} \end{matrix}$

If a Sequence of Moves Yields the Identity then applying those Moves to the Physical Cube Means it is Solved

Suppose that

s = \prod_{i = 1}^{n} m_{i} i n G

, and suppose that

C = do_move (E, move_seq (s))

, then

do_move (C, move_seq (s^{- 1})) = E