The Professor's Cube

IX. Solve the Bottom Inner Edges

The remaining 4 inner-edges are already on the bottom layer, where they belong. The first step is to arrange them in the correct positions. At this point, there are three possibilities:

• NO inner-edge is in place
• Only 1 inner-edge is in place
• All 4 inner-edges are in place

If NO inner-edge is in place, then use the sequence below:

Before

N- B- N+ B2 N- B- N+

After

You only need to do this sequence once.

Note: Ignore the diagrams. As long as you keep the original top face on the top side, this sequence will guarantee that at least ONE bottom inner-edge will land in place afterwards.

If only ONE inner-edge is in place, then rotate the entire puzzle until the fixed inner-edge piece appears on the bottom front. The remaining 3 inner-edges need to be swapped either clockwise or counter-clockwise.

Exchange Clockwise:		Exchange Counter-Clockwise:
N- B+ N+ B2 N- B+ N+		N- B- N+ B2 N- B- N+

You only need to memorize one of the above. For example, if you choose to memorize the "counter-clockwise" sequence, then use it twice to swap the 3 edges clockwise. Once all 4 inner-edges are arranged in place, get ready for the last step: INVERSION.

Inversion

There are 3 different inversion schemes:

1. Invert 2 adjacent inner-edges
2. Invert 2 opposite inner-edges
3. Invert all 4 inner-edges

For each inversion scheme, you must rotate the entire puzzle so that the inverted edges are positioned exactly like the ones in the diagrams, before attempting the sequence of moves!

~~~~ Case #1: Invert two adjacent edges ~~~~
		N- B- N+ B- N- B2 N+ B2 N- B- N+ B- N- B2 N+ B2		Result: The bottom inner-edges are solved. In fact, all the bottom edges are solved.
______________________________
~~~~ Case #2: Invert two opposite edges ~~~~
		N- B- N+ B- N- B2 N+ B2 N- B- N+ B- N- B2 N+ B2		Result: Two adjacent edges are still inverted. Go back to Case #1, do the sequence, and the bottom edges are solved.
______________________________
~~~~ Case #3: Invert all four edges ~~~~
		N- B- N+ B- N- B2 N+ B2 N- B- N+ B- N- B2 N+ B2		Result: Two adjacent edges are still inverted. Go back to Case #1, do the sequence, and the bottom edges are solved.

As it turns out, the same sequence was used throughout all three cases.
Now that all the bottom edges are solved, the only thing left are the
Middle and Bottom Faces.

@ Notation / Top Face
@ Top Corners @ Top Outer-Edges @ Top Inner-Edges
@ Middle Outer-Edges @ Middle Inner-Edges
@ Bottom Corners @ Bottom Outer-Edges @ Bottom Inner-Edges
@ Middle & Bottom Faces

Return to Mathematica