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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_2where 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_2where 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 L2and try again. this proves prop. 3.

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

mike