During this section, you only have to use one sequence of moves.
Only one sequence is used throughout this page, one that has already been used before; so here is the RERUN...
What we want the move to do: | THE MOVE: | What the move really does: |
To swap a single pair of edges on the bottom layer |
Desc. notation:
b- R b-3 R b+
JS notation: |
It swaps the back and left edges on the bottom layer, while swapping the back and right edges on the top layer. |
As you can see, every time a pair of bottom edges are swapped, a pair of top edges also become swapped. Therefore, the sequence must be used again to re-fix the sacrificial top edges. The result? The edges on the bottom layer are swapped TWICE, which is why we had to polarize the cube to even parity before going ahead with this step.
You might be able to figure out how to individually swap the bottom edges until they are fixed; but just in case, a step-by-step outline was added.
SET UP:Rotate the bottom layer so that the 3 edgesto be exchanged are on the back, left and right. In other words, get the "other edge" at the front. |
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Do the move: | Rotate the bottom layer 1/4 of a turn clockwise |
Do the move again: | Result: | |
Desc. Notation: | b- R b-3 R b+ |
b+3 |
b- R b-3 R b+ |
The puzzle is solved! |
JS Notation: | (0,-1)/(0,-3)/ |
(0,3) |
(0,-1)/(0,-3)/ |
SET UP:Rotate the bottom layer so that the 3 edgesto be exchanged are on the back, left and front. In other words, get the "other edge" at the right. |
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Do the move: | Rotate the bottom layer 1/4 of a turn counter-clockwise |
Do the move again: | Result: | |
Desc. Notation: | b- R b-3 R b+ |
b-3 |
b- R b-3 R b+ |
The puzzle is solved! |
JS Notation: | (0,-1)/(0,-3)/ |
(0,-3) |
(0,-1)/(0,-3)/ |
SET UP:Rotate the bottom layer so that two of the edgesthat need to be swapped are at the back and left, while the other two are at the front and right. |
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Do the move: | Rotate the bottom layer 1/2 of a turn around |
Do the move again: | Result: | |
Desc. Notation: | b- R b-3 R b+ |
b6 |
b- R b-3 R b+ |
The puzzle is solved! |
JS Notation: | (0,-1)/(0,-3)/ |
(0,6) |
(0,-1)/(0,-3)/ |
SET UP:None (don't care). |
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Do the move: | Rotate the bottom layer 1/4 of a turn at any direction |
Do the move again: | Result: | |
Desc. Notation: | b- R b-3 R b+ |
b+3 |
b- R b-3 R b+ |
The bottom layer has 3 edges that are still scrambled. If they need to be exchanged clockwise, then go to Case I. If they need to be exchanged counter-clockwise, then go to Case II. |
JS Notation: | (0,-1)/(0,-3)/ |
(0,3) |
(0,-1)/(0,-3)/ |
This sequence does the exact same thing that other sequence did, but with fewer moves. The moves are are not orthogonal, but the top and bottom layer will always remain square:
Swap the back/right edges on the top layer, while swapping the back/left edges on the bottom layer: |
Desc. Notation: | JS Notation: | ||
b- R b-3 R b+ |
(0,-1)/(0,-3)/(1,1)
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And now that the nightmare is finally over, you can live happily ever after.
Desc. Notation | JS Notation | Comments |
b- R t6 R t6 R b+ |
(0,-1)/(6,0)/(6,0)/(0,1) | Fixes the Equator |
b- R t6 b6 R b+ |
(0,-1)/(6,6)/(0,1) |
Swaps the top and bottom layers |
t+ R b-3 R b-3 R b6 R t- |
(1,0)/(0,-3)/(0,-3)/(0,6)/(-1,0) |
Moves up a single corner to the top |
t+ R b-3 R |
(1,0)/(0,-3)/ |
Fixes the top corners |
b- R t+3 R |
(0,-1)/(3,0)/ |
Fixes the bottom corners |
b- R b+ t+ R t- |
(0,-1)/(1,1)/(-1,0) |
Places the first two top edges |
b- R b-3 R b+ |
(0,-1)/(0,-3)/(1,1)/ |
Places the last two top edges |
b- R b-3 R b+ |
(0,-1)/(0,-3)/ |
Fixes the top and bottom edges |
R t-3 b-3 R t-2 b- R |
/(-3,-3)/(-2,-1)/ |
Polarizing move |
b- R t-3 b-3 R b+ |
(0,-1)/(-3,-3)/ |
Polarizing fix |