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- Indices

(8! · 3^8
· 12! · 2^12)
(3 · 2 · 2) = 4.3252 · 10^19 possible arrangements of the Rubik’s cube
- Probability(not so major)

-The number of possible permutations of the squares on a Rubik’s cube seems
daunting. There are 8 corner pieces that can be arranged in 8! ways, each
of which can be arranged in 3 orientations, giving 38 possibilities for each
permutation of the corner pieces. There are 12 edge pieces which can be
arranged in 12! ways. Each edge piece has 2 possible orientations, so each
permutation of edge pieces has 212 arrangements. But in the Rubik’s cube,
only 1
3
of the permutations have the rotations of the corner cubies correct.
Only 1
2
of the permutations have the same edge-flipping orientation as the
original cube, and only 1
2
of these have the correct cubie-rearrangement parity
- Algorithms

-Every plane of the cube can be rotated either clockwise or counterclockwise in relation to the rest of the cube. As you hold the cube in front of you, these planes are identified as front (F), top (T), right (R), left (L), back (B), and down (D). Each letter indicates a 90 degrees clockwise rotation of the corresponding plane. Counterclockwise, or inverted, rotation is indicated by adding an "i" to the letter: Fi, Ri, etc.
- Inequalities

-we can find bounds on the
worst possible number of moves away from the start state with the following
“pidgeonhole” inequality (number of possible outcomes of rearranging must
be greater than or equal to the number of permutations of the cube):
12 · 11n−1 ≥ 4.3252 · 1019
which is solved by n ≥ 19.
- Groups

-By definition, a group G consists of a set of objects and a binary operator,
*, on those objects satisfying the following four conditions:
• The operation * is closed, so for any group elements h and g in G, h ∗ g
is also in G.

###
- Theorems

-Try repeating the move FFRR on a solved cube until you get back to the
starting position. How many times did you repeat it?(hopefully 6) No matter
what, that number, the size of the subgroup generated by FFRR, must be a
divisor of (8!·3
8
·12!·2
12)
(3·2·2) . Lagrange’s Theorem tells us why.
Before we prove Lagrange’s Theorem, we define a coset and note some
properties of cosets. If G is a group and H is a subgroup of G, then for
an element g of G:• gH = {gh : h ∈ H} is a left coset of H in G.

• Hg = {gh : h ∈ H} is a right coset of H in G.

• The operation * is associative, so for any elements f, g, and h, (f ∗ g) ∗ h = f ∗ (g ∗ h).

• There is an identity element e ∈ G such that e ∗ g = g ∗ e = g.

• Every element in G has an inverse g −1 relative to the operation * such that g ∗ g −1 = g −1 ∗ g = e.

###
- Parity

-Permutations can also be described in terms of their parity. Any length n
cycle of a permutation can be expressed as the product of 2-cycles.2 To
convince yourself that this is true, look at the following examples: (1 2) = (1 2) (1 2 3) = (1 2)(1 3) (1 2 3 4) = (1 2)(1 3)(1 4) (1 2 3 4 5) = (1 2)(1 3)(1 4)(1 5) The pattern continues for any length cycle.

###

- Cayley Graphs

-A useful way to gain insight into the structure of groups and subgroups is
the Cayley graph. The following properties describe a Cayley graph of a
group G:
• Each g ∈ G is a vertex.
• Each group generator s ∈ S is assigned a color cs.
• For any g ∈ G, s ∈ S, the elements corresponding to g and gs are joined
by a directed edge of color csThere are so many more but some are to complex to be understood and our level.