siva.shanmukh
Member
This is something very frequently used in FMC.
Apply scramble on solved cube, solve till a point. Invert the whole thing and work from there. Something like this.
If the problem statement is S,
On identity(solved cube) execute S M1 (till one can't figure a good path from there on)
On identity(solved cube) execute M1' S' M2 (till one can't figure a good path from there on)
.
.
.repeat this till Ma S Mb (or the inverse version) is Identity (solved cube) and call Mb Ma as solution.
I understand how both S M1 and M1' S' has the same set of solved cubies, but what I don't understand is how they have same number of solved blocks
Eg: (from FMC, weekly 2012-06)
S = B D2 R U B D' R D2 F U D' B2 D' L2 D' F2 R2 B2 D2 R2
M1 = B2
Now S M1 only has two paired blocks not in their correct position. I don't understand how it so happens that M1' S' has two paired blocks (Not the same ones as in S M1) in different relative positions.
Can someone please explain?
Apply scramble on solved cube, solve till a point. Invert the whole thing and work from there. Something like this.
If the problem statement is S,
On identity(solved cube) execute S M1 (till one can't figure a good path from there on)
On identity(solved cube) execute M1' S' M2 (till one can't figure a good path from there on)
.
.
.repeat this till Ma S Mb (or the inverse version) is Identity (solved cube) and call Mb Ma as solution.
I understand how both S M1 and M1' S' has the same set of solved cubies, but what I don't understand is how they have same number of solved blocks
Eg: (from FMC, weekly 2012-06)
S = B D2 R U B D' R D2 F U D' B2 D' L2 D' F2 R2 B2 D2 R2
M1 = B2
Now S M1 only has two paired blocks not in their correct position. I don't understand how it so happens that M1' S' has two paired blocks (Not the same ones as in S M1) in different relative positions.
Can someone please explain?