Date: Sat, 14 Jan 95 17:18:30 -0500
From: michael reid <mreid@ptc.com >
~~~ ~~~ Subject: more on superflip

recently i said:

when searching for superflip in the face turn metric, it's
sufficient to search through depth 17 in stage 1!

since posting this, i've realized that we can do much better.
here's my current approach. everything below refers to the
face turn metric. (i have similar reductions for quarter turns,
but they're not quite as good.)

proposition 1. there is a minimal sequence for superflip of the form

sequence_1 sequence_2

```where  sequence_1  is in stage 1,  sequence_2  is in
stage 2, and  sequence_1  is at most  17f  long.
```

proof. consider the different possibilities for the length of a
minimal sequence for superflip: 20f, 19f, 18f, 17f or less.
in the first case, we already know a maneuver of the form.
in the second case, my discussion on thursday shows that
we'll have such a maneuver. in the case of 18f , we may
suppose that the last face turned is U , so we'll have such
a maneuver. and in the last case, we may take sequence_2
to be the empty sequence. this proves prop. 1.

proposition 2. there is a minimal sequence for superflip of the form

```R1  sequence_1  sequence_2
```
```where  sequence_1  is in stage 1,  sequence_2  is in
stage 2, and  sequence_1  is at most  16f  long.
```
```proof.  consider the maneuver given by prop. 1.  by applying one of
the 16 symmetries that fix the  U - D axis, we may suppose
that the first turn of  sequence_1  is either  U1, U2, R1,
or  R2.  in the case of
```
```U1  sequence_1  sequence_2,
```

replace this by

```sequence_1  sequence_2  U1,
```

and try again. handle the cases starting with U2 and R2
similarly. we will either exhaust the stage 1 part of the
sequence (which is impossible, since superflip isn't in
the subgroup of stage 2) or we'll wind up with a manuever
starting with R1 , as desired. this proves prop. 2.

there's still some more symmetry left to exploit.

proposition 3. there is a minimal sequence for superflip of one of the
forms

```R1 F1  sequence_1  sequence_2,
R1 F2  sequence_1  sequence_2,
R1 F3  sequence_1  sequence_2,
R1 U1  sequence_1  sequence_2,
R1 U2  sequence_1  sequence_2,
R1 U3  sequence_1  sequence_2,
R1 L1  sequence_1  sequence_2,   or
R1 L3  sequence_1  sequence_2,
```
```where  sequence_1  is in stage 1,  sequence_2  is in
stage 2, and  sequence_1  is at most  15f  long.
```
```proof.  by applying the symmetry  C_R2  if necessary, we may suppose
that the second turn of the maneuver given by prop. 2 is one
of  F1, F2, F3, U1, U2, U3, L1, L2  or  L3.  this gives nine
cases.  in the case
```
```R1 L2  sequence_1  sequence_2,
```

replace this by

```R1  sequence_1  sequence_2  L2
```

and try again. this proves prop. 3.

i have these cases running right now, and i hope to have results soon!

mike