Date: Tue, 17 Dec 91 16:18:16 -0500 (EST)
From: Dan Hoey <hoey@aic.nrl.navy.mil >
~~~ 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
Hoey@AIC.NRL.Navy.Mil