by Richard Taylor : 2026-02-23
For my last two FMC competitions I have been using domino reduction (DR) and half-turn reduction (HTR). After HTR the cube can be solved with only L2, R2, F2, B2, U2, D2 moves.
That should be doable, but it's not always as easy as it seems. So there is another step, called floppy reduction (FR), which may help by reducing the cube to only 4 of those moves - eliminating either L2/R2 or F2/B2 or U2/D2.
When I looked at this I wasn't sure it was worth the effort though, as simply solving the corners and then using edge insertions seemed to often give a better result than a direct half-turn finish.
However, in my last competition, one of my solves threw up a situation where FR could have improved my solution. After getting to HTR, solving the corners and then solving the edges with insertions I noticed that the last 13 moves of my solution were :-
L F2 B2 U2 F2 D2 B2 L2 D2 F2 R2 D2 L2
I felt that those last 12 half-turns might be sub-optimal, so I scrambled a cube with the inverse L2 D2 R2 F2 D2 L2 B2 D2 F2 U2 B2 F2 and tried to solve it in less than 12 moves.
Unfortunately in the competition I failed, but when I got home I used a computer to find the optimal solutions and saw it could be done in 10 moves. What's more, some of the solutions started with L2 which cancels a move with the L just before HTR is achieved.
F2 B2 U2 F2 D2 B2 L2 D2 F2 R2 D2 L2
=
L2 U2 R2 U2 L2 B2 D2 F2 D2 F2
So that's only 10 moves instead of 13. And looking at the new set of half-turns it is a floppy-reduction :-
L' // HTR
U2 R2 U2 L2 // FR
B2 D2 F2 D2 F2 // finish
L2/R2 is the axis we are reducing. The set-up is U2 R2 U2 L2 and after that we can solve the cube with only F2, B2, U2, D2 moves ... in this case B2 D2 F2 D2 F2 we don't even need all 4 of them.
I think in general you will often find that after HTR you can find a shorter solution using insertions than you will get with floppy reduction (FR). But there are cases when FR is optimal and cases where, even after solving the cube with insertions, you can improve your complete solution by shortening a sequence of half-turns using FR.
So I wish I'd paid a bit more attention to floppy reduction before this competition. I still might not have been able to find the optimal FR, but if I had then it would have turned my PR of 31 into a 28.