Permutations and the Facelet Level 

If we look at clean the cube, we see 6*9 facelets. 

If we apply a move to the cube, the facelets are rearranged. Such a rearrangement
is called a permutation. If we apply for example a Fmove to the cube depicted above we get the following result:


To explain the representation of such permutations, we will only look at the yellow facelets for a moment. There are two possibilities for this representation in the example.
Because the first row of the tables is always the same, we can omit this row. So we can write just (F3,F6,F9,F2,F5,F8,F1,F4,F7) in the is carried to representation or (F7,F4,F1,F8,F5,F2,F9,F6,F3) in the is replaced by representation. In most cases we will not use the short form without a table here for the sake of clearness. We use the first representation on the facelet level, and the second
on the cubie level. 

We are able to define a product of two permutations. For example
*
=
because for example F1›F2 by the first permutation and F2›F6 by the second, so we have F1›F6 in the product. 

The multiplication of permutation has some similarities with the common multiplication with numbers, but there is one big difference: While for example 3*5=5*3, you usually may not exchange the order of the two permutations. But in the above example, we have
*
=
and this is something different. The multiplication of permutations is not commutative. 

Another important term is the inverse permutation. Consider the Fmove
and the permutation
In this case
*
=
This last permutation does nothing at all, so when we multiply a permutation
with its inverse, we get the identity permutation
I. In fact, in this example the second permutation is the representant
of F', so we have F*F' = I, which is quite obvious. 

In the file CubeDefs.htm you can see the full definition of the basic moves. For example F:=(U1,U2,U3,U4,U5,U6,R1,R4,R7,D3,R2,R3,D2,R5,R6,D1,R8,R9,F3,F6,F9,F2,F5,F8,F1,F4,F7, written in the short form without a table. Not only the moves can be viewed as a permutations, every scrambled cubed can be written as a permutation. Because U1›R3, U2›L2, U3›D3, U4›U8,... this cube has the representation (R3,L2,D3,U8,...). If you solve this cube, you in fact will try to find the inverse permutation of this cube composed as a product of the permutations corresponding to the elementary moves U,U2,U',R,R2,R'..... Recall that the product of a permutation with the inverse gives the identity permutation, and this is the clean cube. The solving algorithm of Cube Explorer tries to find short products for this inverse. For the example in the picture above it finds in a few seconds R2*L*U2*L2*D*R2*U2*L'*D2*R'*U*B*R'*F2*L2*B2*L2*B2 But representing permutations on the facelet level is not effective for a fast computation. There are two more levels to cope. 
