[next] [prev] [up] Date: Tue, 17 Dec 91 16:18:16 -0500 (EST)
[next] [prev] [up] From: Dan Hoey <hoey@aic.nrl.navy.mil >
[next] ~~~ [up] Subject: Re: Rubik's cube dice tops (Spoiler)

Last week ronnie@cisco.com (Ronnie Kon) challenged us to find Rubik's
cube patterns with dice pips for 1, 2, and 3 on the three pairs of
opposite sides. He claimed it could be done in fourteen HST, where
one HST is a turn of a face or center slice by 90 or 180 degrees. I
responded that it could be done in thirteen HST. Here is how. I will
use this opportunity to practice the enhanced Varga Rubiksong I
described (unfortunately with many typos) on 22 Feb 90.

The (only such) pattern is the composition of Four-Spot and Laughter.
We have long known the processes

        ris-fos tis-fos, or (RL)^2 FB' (TD)^2 FB', for Four-Spot and
fon-ron fon-ron fon-ron, or (FBRL)^3,              for Laughter.

When we compose them, the F and B moves combine and cancel to produce

ris-fos tis-fi ron-fon ron-fon ron, or (RL)^2 FB' (TD)^2 F^2 (RLFB)^2 RL.

This 14 HST process is presumably something like what Ronnie had in
mind. But since this pattern commutes with ris, or (RL)^2, we can get
the same pattern with the conjugate process

fos tis-fi ron-fon ron-fon ran, or FB' (TD)^2 F^2 (RLFB)^2 R'L'.

This uses only 13 HST. This is also the shortest process I know of in
the normal metric: 18 QT, which is not so bad for the combination of
two 12 QT processes.

I suggested that perhaps 12 HST would be sufficient, but I have not
found such an improvement. Nor do I know whether 13 HST is the best
that can be done: it seems that proving 13 HST optimal would require
examining about 160 million positions, almost as many as the 200
million it would take to prove 18 QT optimal.

Dan Hoey

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