# Maneuver and its Inverse (question)

Discussion in 'Puzzle Theory' started by siva.shanmukh, Feb 13, 2012.

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1. ### siva.shanmukhMember

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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.

2. ### qqwrefMember

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Suppose some sequence P starts from solved, and ends up with the pieces at UR and URB are paired up (just as an example). Then that means that those particular two pieces start out paired (since we start at solved) and also end up paired, but possibly in a different position. We can also say that if we apply P to any position, then if the two pieces in the original position of our pair start out paired, then they will end paired in UR/URB - and vice versa.

This means that if you execute P starting from the scramble P', so that we end up at solved, some two pieces (specifically the two pieces that are in positions UR/URB at the end) will be paired at the end whenever they were paired at the start. Since we KNOW they are paired at the end (the cube is solved, after all), they must also be paired at the start, that is, in scramble P'. So the scramble P' has a pair, and it contains the UR and URB pieces.

You can see from this approach that we can do it to track any number of pairs at once. So if the scramble P has two pairs, then we know P' has two pairs (and we can predict which pieces they will be from where the pairs in P are).

3. ### siva.shanmukhMember

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I totally get it now.. Not just multiple pairs, but all sorts of vague groups do re-appear in the way you explain on doing the reverse maneuver. That was really well explained.

Thanks a lot.

4. ### CieloMember

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S M1 has a block A, then there exists a sequence X (exactly, X is some rotation of the cube), and block A is among the solved cubies of S M1 X.
As you posted, block A is among the solved cubies of X' M1' S', too. Since X' is a rotation, M1' S' has a block B.

How would you use this in a fmc solve?

6. ### siva.shanmukhMember

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This I already knew. I mean if there is a set of solved cubies, they would be solved in the inverse maneuver as well. My question is regarding unsolved blocks. anyway don't bother, I got my answer from qqwref's reply

In FMC, if you are stuck at a point, say right after solving a 2x2x2 block (by saying stuck, I mean not getting any optimal maneuver from there on), you could just invert the scramble and the partial solution you have got so far on a solved cube and try to proceed from there. This not only gives a different unsolved state with the same solved block, but also lets you use it as your solution. for example if Ma S Mb renders the cube solved where S is your scramble, then Mb Ma is a solution for S. You can prove it in the following way.

Ma S Mb = I
Ma' Ma S Mb Ma = Ma' I Ma
S Mb Ma = I

Hence Mb Ma is a solution for S

Last edited by a moderator: Feb 14, 2012
7. ### CieloMember

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My answer is regarding unsolved blocks too. I meaned to answer you question using what you have known in #1

What I said is that we can extend the cube group by adding rotations x,y,z. Then for a unsolved block A of S M1, we can rotate the whole cube (the rotation is X) so that we put the block A in its initial position. In this way the cubies of block A become solved cubies of S M1 X.

8. ### siva.shanmukhMember

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If a set of cubies are not solved in a S M1, how can they be solved in S M1 X if X is a cube rotation? No matter how much you rotate a cube, it would still be in the same state. I mean S M1 X would represent the same state of the cube as S M1 if X is just a cube rotation.

9. ### CieloMember

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You are right. I didn't explain clearly.

Maybe we can see the cube from a different point of view. That is, the center cubies can change positions (by rotation), just like edges and corners.
Then "a block is solved" means the block is in the same position as it was in the initial state. For example, the cubies in R layer is solved in R x', and other cubies are not solved.

10. ### siva.shanmukhMember

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In that case, your explanation still doesn't help. If you want to see for yourself, you could execute the example I posted in the first post of the thread.

11. ### CieloMember

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S = B D2 R U B D' R D2 F U D' B2 D' L2 D' F2 R2 B2 D2 R2
M1 = B2
S M1 has two blocks: DB&DBL (now in RU&RUB), LB&LBU (now in UL&ULB).
For the first one, X = z y', so after executing S M1 X, DB&DBL are "solved". As a result, DB&DBL are also "solved" if we execute X' M1' S'.
In fact, DB&DBL block is in RU&RUB after X'. So M1' S' moves the RU&RUB block to DB&DBL.

After writing down this, I find my explanation was too complicated -_-| But It still can answer your question in the first post.
Just as qqwref posted: S moves a block from position A to B, then S' moves a block from position B to A.

12. ### siva.shanmukhMember

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