[next] [prev] [up] Date: Tue, 24 Jan 95 23:58:00 WET
[next] [prev] [up] From: Martin Schoenert <Martin.Schoenert@math.rwth-aachen.de >
[next] [prev] [up] Subject: Re: Re: Re: Models for the Cube
Jerry Bryan wrote in his message of 1995/01/18

It is well known that G[E] must have an equal number of even and
odd permutations. If we generate G[E] as <Q[E]>, it is also the case
that there are just as many positions an even distance from Start as
an odd distance from Start because the parity of the distance from
Start is the same as the parity of the permutation if we restrict
ourselves to quarter turns.

If you view the elements of G[E] as permutations on 24 facelets, then all
elements are even. But if you forget for the moment the orientations of
the edges, and view each element as only permuting the 12 edges, then
there is an equal number of even and odd elements in G[E]. And then,
since each quarter face turn cyclically permutes four edges, there must
indeed be just as many states an even distance from Start as there are
states an odd distance from Start.

Jerry continued

But in the computer search for God's Algorithm for edges only cubes,
there were not equal numbers of positions an even distance from Start
as an odd distance from Start. The computer search used the coset model
G[E]/C[E], where this notation means the set of cosets of C, *not* the
factor group. In and of itself, the mismatch between the number of
positions at an even distance from Start and at an odd distance from
Start should not pose a problem. It is not clear to me what it means
to speak of the "parity" of a coset of C, and half of each coset will
be even and the other half will be odd. Hence, it is not clear that
a particular coset must *a priori* be an odd or even distance from

Allow me to translate this into the language I introduced in my other
message. G[E] is the model for the basic puzzle. C[E] is the subgroup
of essentially free elements. We consider two elements of G[E] to be
essentially equal if they lie in the same left coset of C[E] in G[E].
The cost of an element <g> of G[E] (or the distance from the Start), is
the length of a shortest process effecting <g>, where we only count the
quarter face turns, not the rotations of the rigid cube. It is clear
that two essentially equal elements have equal costs.

Jerry continued

But if we map each coset to an element of G[E], it is meaningful to
speak of the parity of the element of G[E]. And if the elements of
G[E] to which we map the cosets form a subgroup, then there must be an
equal number of odd and even elements in the subgroup
(unless they are all even?!). If the mapping respects
costs in the sense of maintaining the same distance from Start, then
I don't understand how we can avoid a conflict between the equal
number of even and odd permutations in the subgroup of G[E] and
the unequal number of even and odd distances from Start in the coset
model G[E]/C[E].

Pick one edge, say the right upper edge. Then each coset of C[E]
contains one representative that fixes this edge. The set of those
representative is a subgroup U, generated by L[E],D[E],F[E],B[E].
Formally U is a supplement for C[E] in G[E]. It is a model for the
essential edges only cube. And indeed it contains the same number
of even and odd permutations. So far so good.

But we must now be carefull how we measure the cost of elements in U.

If we measure by taking the length of a shortest process effecting such
an element <u> in U using only the generators L[E],D[E],F[E],B[E] (and
their inverses), then some elements will appear more expensive than they
really are. For example r[E]'*R[E] should have cost 1, but there is no
process of length 1 effecting this element using only the generators
above. So we must take a larger generating system. As I described in my
other message, the generating system to take is the set of all elements
in U that should have cost 1. This gives us the generating system
{ L[E], D[E], F[E], B[E], r[E]'*R[E], u[E]'*U[E], L[E]', ... }.
Still, so far so good.

So where is the problem? Well the new generators r[E]'*R[E] and
u[E]'*U[E] are *even* permutations. And since not all generators
are odd permutations any more, the argument that the number of
elements with even cost and the number of elements with odd cost
must be equal, simply doesn't work anymore.

Have a nice day.


-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

[next] [prev] [up] [top] [help]