Message #3733

From: Marc Ringuette <ringuette@solarmirror.com>
Subject: 3-cycle demo on phys 2^4
Date: Fri, 23 Jun 2017 17:26:55 -0700

Hi, 4D cubers,

Today I ticked off another item on my to-do list for Melinda’s physical
2^4 puzzle: I found a 3-cycle of corners using only the "common subset"
moves** on the phys 2^4.

LF'  (FO2 LO' FO2 LO FO2) LF  OR2<br>
LF'  (FO2 LO' FO2 LO FO2) LF  OR2

It cycles LUFI to RUFO to RDBO, and consists of 16 twists, each of which
affects half the puzzle.

Video demo on the phys 2^4 (90 secs): http://youtu.be/S8hm3CupPJA

In the video description is a link to my vid replicating the 3-cycle on
mc4d using the same colors. Or, here is the mc4d sequence in log file
format:
103,1,1 79,-1,1 79,-1,1 101,-1,1 79,-1,1 79,-1,1 101,1,1 79,-1,1 79,-1,1
103,-1,1
133,-1,1 133,-1,1 133,-1,1 133,-1,1 23,-1,2 23,-1,2 103,1,1 79,-1,1
79,-1,1 101,-1,1
79,-1,1 79,-1,1 101,1,1 79,-1,1 79,-1,1 103,-1,1 23,-1,2 23,-1,2.

** Note 1 - the meaning of "common subset": In the phys 2^4, we forbid
moves that can’t be done in one click on mc4d (restacks, clamshells, and
4-piece subface twists). In mc4d, we forbid moves that can’t be done
in one simple twist on the phys 2^4 (i.e. 90 degree twists or rotations
that do not hold the R-L axis fixed; all half turns are OK, even those
without an R or L as the first or second letter of their name).

** Note 2: I am 99% certain that all moves in both puzzles have either
simple moves or macros (sequences of legal moves) in the other puzzle.
The "common subset" is a proper subset of the moves available in each
puzzle, that is useful for direct translations of algorithms from one to
the other. It limits the macro/sequence length to 1 in both directions.
Algorithms not in the common subset take a greater number of moves
in one puzzle or the other.

** Note 3: The common subset cannot solve the puzzle because it cannot
fully orient each piece. It can’t move stickers on or off of the
combined R+L faces. On the phys 2^4, we must add either FOro or FUro
(or something equivalent) to the common subset in order to fully solve
all 2^4 scrambles. I demoed those in my video #3 on June 10th.

Cheers
Marc