From:

~~~ ~~~ Subject:

A couple of days ago, I said that proofs are a good idea. I'll say it

again today with a redder face.

Yesterday I discussed the edge group positions

I = Solved,

P = Pons Asinorum (or Mirror),

E = All edges flipped, and

PE = P E = Pons Asinorum with all edges flipped

and the function from the edge group to 4-tuples of distances

f(X)=(d(I,X), d(E,X), d(P,X), d(PE,X)).

I wrote:

?? If f(X)=(a,b,c,d), then conjugation shows us that ?? ?? f(X E)=(b,a,d,c), f(X P)=(c,d,a,b), and F(X PE)=(d,c,b,a). ?? ?? So the set of quadruples has the symmetries of the rectangle. ??

The first sentence is incorrect, though the argument as a whole is

reparable.

First, I'll do what I should have done yesterday, and define the

distance function d(X,Y). We want the minimum length process Z such

that X Z = Y. But premultiplying both sides by X', we have Z = X' Y.

So I define d(X,Y)=Length(X' Y). From the properties of the length

function (Length(I)=0, Length(X)=Length(X'), and

Length(X Y)<=Length(X) + Length(Y)) we can conclude that d(X,Y) is a

metric.

Suppose f(X)=(a,b,c,d). I claim f(E X)=(b,a,d,c), f(P X)=(c,d,a,b), and F(PE X)=(d,c,b,a).

Proof: To show f(E X)=(b,a,d,c), first observe that I=I', E=E', and

P E = E P.

d(I,E X) = Length(I' E X) = Length(E' X) = d(E,X), so d(E,E X) = d(I, E E X) = d(I,X); d(P,E X) = Length(P' E X) = Length((PE)' X) = d(PE,X) so d(PE,E X)=d(P,E E X)=d(P,X).

To show that f(P X)=(c,d,a,b), exchange P and E in the above argument.

To show that f(PE X)=(d,c,b,a), use both occurrences of the argument.

QED.

So the idea of yesterday's message is correct, but I had X E, X P, and

X PE instead of E X, P X, and PE X, respectively. I would show you a

counterexample to yesterday's formulation, but it turns out there is

none. I claim that f(X,E)=f(E,X), f(X,P)=f(P,X), and f(X,PE)=f(PE,X).

Proof: Recall that E commutes with every element of the Rubik cube

group, so f(X E)=f(E X). It turns out that ``up to M-conjugacy'', P

commutes with every element of the edge group as well. For P performs

a mirror-reflection of the edges, and so can be regarded as an element

of M acting on the edge group. So P' X P = Xbar is an M-conjugate of

X, and X P = P Xbar. Since Length(X) agrees on M-conjugates, so does

d(X,Y), and so f(X), so f(X P)=f(P Xbar) = f(P X). Finally,

f(X PE) = f(X P E) = f(E X P) = f(P E X) = f(PE X). QED.

So it turns out it that the statement about f was true. But I am no

less embarrassed for asserting it, for I had no reason to think it

would be true. It's only rescued by the surprising commutativity of

the Pons Asinorum.

Finally, I would like to note something that I nearly included in

yesterday's message, but yanked when I decided it was false:

f(X')=f(X). Now I'll prove it:

Proof: For W among {I,E,P,PE}, we have X W = W Xbar, for Xbar an

M-conjugate of X. So

d(X,W)=Length(X'W)=Length(W'Xbar')=Length(W'X')=d(W,X').

QED.

Dan Hoey

Hoey@AIC.NRL.Navy.Mil