).
I say that a process *solves* a basic stage , if

= .
Assume that G and S are such that for each basic state there is a
process

that solves (technically this means that the group of
operations on G operates transitively on the set G of basic states).
The goal of the puzzle is to find such a process for each basic state.
If a process

solves a basic state , then the inverse process

',
that we get by reversing the sequence and replacing each simple operation
by its inverse, transforms into , and I say

' *effects* .
Finally assume that G and S are such that if processes and
effect the same basic state , then they induce the same operations,
i.e., then = for any other basic state in G
(technically this means that the group of operations on G operates
regularly on the set G of basic states).
This then allows us to identify the set G of basic states and the group
of operations on G. This in turn allows us to view G as a group, with
the generating system S.
Sometimes it is necessary to distuingish between a process, the operation
it induces, and the basic state it effects. When such a distinction is
unnecessary, I shall simply talk about the *element* of G.
I call the group G a model for the *basic puzzle*.
For an example take Rubik's cube. There are 24 * 43252003274489856000
basic states. The simple operations are the face turns (or quarter face
turns if you prefer) and the rotations of the rigid cube. Obviously each
state can be solved and it is easy to verify that two processes that
effect the same basic state induce the same operation. The group CG is
the model for the basic Rubik's cube.
The Essential Puzzle
--------------------
Next we need to add the notion that, while all basic elements (states)
are different, some are more different than others.
Assume that there is a subgroup of *essentially free elements* F.
Assume that we shall consider two elements and to be
*essentially equal*, if there is an essentially free element in F
such that = . Then the sets of essentially equal
elements are of course exactely the (left) cosets ( F). We shall
call such a coset an *essential element*.
Assume that we have selected for each coset ( F) a representative
r( F). Define the operation '*' on the set of left cosets by
( F) * ( F) := (r( F) r( F) F), i.e., the product of
two cosets is the coset containing the product of the representatives.
If the set of essential elements with this operations is a group,
then I call this group a model for the *essential puzzle*.
Note that there is no guarantee that we can select the representatives
such that '*' defines a group. That is, for some puzzles there may not
be a group model for the essential puzzle. This for example happens if
G = < (1,2), (1,2,3,4) > is the symmetric group on four points and
F = < (1,2)(3,4) >.
Furthermore there is no guarantee that each selection that gives us a
group, gives us the same group. That is, for some puzzles there may be
different nonisomorphic group models for the essential puzzle. This for
example happens if G = < (1,2), (1,2,3,4) > and F = < (1,2), (1,2,3) >
is the stabilizer of one point, in which case C4 = < (1,2,3,4) > and
V4 = < (1,2)(3,4), (1,3)(2,4) > are models.
But if F is a normal subgroup, then it doesn't matter how we select the
representatives; the operation '*' always gives us the factor group.
Also if there is a supplement E of F (i.e., a subgroup E, such that
G = E F and E F = { }), then selecting the elements
of E as the representatives of the cosets gives us a group model for the
essential puzzle. This group model is of course isomorphic to E
(but note that there can be nonisomorphic supplements).
But there can be group models for the essential puzzle that come neither
from the factor group nor from a supplement. This for example happens if
G = < (1,2,3,4,5,6,7,8), (2,8)(3,7)(4,6) > is the dihedral group of size
16 and F = < (1,2)(3,8)(4,7)(5,6), (1,5)(2,6)(3,7)(4,8) >, in which case
there are again two nonisomorphic models.
For example in the case of Rubik's cube we usually consider two basic
states to be equal if they are related by a rotation of the rigid cube.
That is the subgroup F of essentially free elements is the subgroup C of
rotations of the rigid cube in this case. The usual group model for the
essential Rubik's cube is the supplement G of C.
The Costs
---------
Finally we need the notion of costs, both for the basic puzzle and
the essential puzzle.
In general let G be an arbitrary group, X a generating system for G that
is closed under taking inverses (that is, for each in X, ^-1 is
also in X), and Z a subgroup of G. Then roughly the cost of an element
in G wrt. X and Z is the length of a shortest process effecting ,
where we only count the generators in X, not the terms in Z.
More formally define G_(0) := Z and G_(l+1) := G_(l) (G_(l) X Z).
Since X is a generating system, there is a finite d such that G = G_(d)
(and the smallest such d is called the diameter of G wrt. X and Z).
We say that elements which lie in G_(l) but not in G_(l-1) have *cost* l.
For the basic puzzle group G, we obviously use the cost function
cost_G for G wrt. the generating system S and the subgroup F.
So the cost of a basic state is the length of a shortest process
effecting , where we count only the simple operations ~~ in S,
not the elements for the essentially solved operations in F. Note
that of course the costs of two essentially equal elements are equal.
For the essential puzzle group E, we want to find a cost function
cost_E that preserves this cost. That is, we want
cost_E( F ) = cost_G( ). So the question is, can we
choose a generating system X_E of E and a subgroup Z_E of E such that
the cost function for E wrt. to X_E and Z_E has the above property.
If you think about it for a moment, you will see, that we actually
don't have any choice if we require cost_E( F ) = cost_G( ).
Namely Z_E is simply the subgroup of E of the elements with cost 0.
But if cost_E( F ) is 1, then cost_G( ) must be 1,
so must be in F, and then ( F) is F, i.e., the identity of E.
Likewise X_E is simply the subset of E of the elements with cost 1.
But if cost_E( F ) is 1, then cost_G( ) must be 1,
so must have the form ~~~~ , so we see that X_E must be
the set { ( ~~~~ F) | in F, ~~~~ in S }.
Now it turns out that those two conditions are not only necessary, they
are in fact sufficient. That is, if cost_E is the cost function for E
wrt. the generating system X_E = { ( ~~~~ F) | in F, ~~~~ in S }
and the subgroup Z_E = { ( F) }, then cost_E preserves the cost,
i.e. cost_E( F ) = cost_G( ).
So for example in the case of Rubik's cube the cost of an element of
CG is the length of shortest process effecting , where we only count
the face turns (or only the quarter face turns), not the rotations of the
rigid cube. A model for the essential Rubik's cube is the supplement G
generated by the face turns (without the rotations). Because for each
process we can collect all the rotations of the rigid cube at the end of
a process, we see that the set X_E is simply the set of the face turns.
For another example take the edges only cube CG[E]. The cost of an
element is again the length of the shortest process effecting ,
where we only count the face turns, not the rotations. A model for the
essential edges only cube is E = (i.e., a subgroup
of CG[E] that fixes one edge) (Confusion warning: running out of letters
again, the first E stands for *e*ssential, the others for *e*dges). If
we want the cost function for E to preserve the costs in CG[E], we must
take the generating system X_E = { ( ~~~~ F) | in F, ~~~~ in S } =
{ L[E],D[E],F[E],B[E],r[E]'*R[E],u[E]'*U[E], L[E]', ... }. Otherwise
some elements of E, will appear more expensive than they really are.
Summary
-------
Assume we have a puzzle modelled by a group G of basic elements with
a generating system S of simple elements and their inverses.
Assume that we have a subgroup F of essentially free elements, and that
we call two elements essentially equal if the lie in the same left coset
of F in G.
Given a system of representatives for the cosets of F in G, we define
the product of two cosets as the coset containing the product of the
representatives of the two cosets. If this multiplication turns G/F
into a group, we call this group a model for the essential puzzle.
Note that such a model need not exist, i.e., it may happen that no system
of representatives gives a group. Also such a model need not be unique,
i.e., different systems of representatives may give nonisomorphic models.
The cost of an element in G is the length of a shortest process effecting
this element, where we count only the simple elements from S, not the
essentially free elements from F.
Then the cost of an element in G is equal to the cost of its left coset
in G/F wrt. the generating system { ( ~~~~ F) | in F, ~~~~ in S }.
Have a nice day.
Martin.
PS. I admit this more than a *bit* formal and long.
Count it as my submission for the understatement-of-the-year award ;-)
-- .- .-. - .. -. .-.. --- ...- . ... .- -. -. .. -.- .-
Martin Sch"onert, Martin.Schoenert@Math.RWTH-Aachen.DE, +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany
~~