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Intuition Competition [Weekly]

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elrog

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I am terribly sorry. I mistyped the algorithm.

It should be: y' (L' U' L U) (R U R' U') (U' L' U L) (U R U' R')

I will correct it in my post.
 

Dane man

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Current standings. These are not the final results.

1st - elrog with 15 moves
2nd - cmhardw with 12 moves. The solution works and there was a wonderfully clear explanation, but used algorithms that were already known. Was simply moved down instead of DQ'd because he explained why the algorithm works.

HM - Lucas Garron with 12, 13, and 18 moves. The solution works, but there was no sufficient explanation of the origin or workings of any algorithm. In doing so, hinted that the solution of 11 moves was "popular" and therefore already known. Given an honorable mention because clearly understood what the algorithms did and why, despite the lack of sufficient explanation. (If he can give a sufficient explanation of these algorithms, then the rankings will be adjusted to include this entry).

Final standings and new challenge coming soon!
 
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Current standings. These are not the final results.

1st - elrog with 15 moves
2nd - cmhardw with 12 moves. The solution works and there was a wonderfully clear explanation, but used algorithms that were already known. Was simply moved down instead of DQ'd because he explained why the algorithm works.

HM - Lucas Garron with 12, 13, and 18 moves. The solution works, but there was no sufficient explanation of the origin or workings of any algorithm. In doing so, hinted that the solution of 11 moves was "popular" and therefore already known. Given an honorable mention because clearly understood what the algorithms did and why, despite the lack of sufficient explanation. (If he can give a sufficient explanation of these algorithms, then the rankings will be adjusted to include this entry).

Final standings and new challenge coming soon!

what about me?
 

elrog

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Just because it is a commutator does not mean you understand it. Commutators are algorithms as well. You have to show that you do understand. That's the point in this.
 

Dane man

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New challenge!

You must create the following pattern from this angle on the cube. It doesn't matter what the rest of the cube looks like.
image.jpg

Yeah yeah, patriotic I know. Have fun!

PS: If you're not particularly fond of this color scheme, you may choose your own, but it must achieve this pattern.
 

Lucas Garron

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Hmm. I can't use Stefan's alg, then. Here's something very simple:

We're going to swap four pieces around UBR with FDR, and then do the same with FDR and LBD

Setup onto U: U' R D R F' D' R'
Swap with U2. Full commutator: [x y', [U' R D R F' D' R': U2]]

Flip a corner on D: R U2 R' F' U2 F
Full commutator to flip rotate corners: [R U2 R' F' U2 F, D]
Conjugated to handle opposite corners with a cancellation: [R U2 R' F' U2 F, D]

Invert both of the parts, and rotate the cube to align the color scheme:
y2
[x' R2: [D, R U2 R' F' U2 F]]
[[U' R D R F' D' R': U2], x y']
y'


It's easy to take off another 2-3 moves using the right rotations, but I'm obviously not going to movecount here.

EDIT: There's also an obvious setup into 3-cycles of pairs around the rotated ring:

Setup to move DFR next to RUB: F'
Commutator to swap pairs: [U L2 U', r']]
Repeat twice and combine with corner twist from previous alg: y2[x' R2: [D, R U2 R' F' U2 F]][F': [U L2 U', r']][D: [R' B2 R, f]]y'
 
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elrog

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I started off holding blue on top, red on right, and white on front.

First off, I started this challenge trying to solve the corners as they were harder. The corners have only 1 position they can be where as the edges have multiple edges not shown. The corners also require you to orient, but not permute, them.

I realized that I could swap the URF corner with the DLB corner and then swap them back into the correct orientation.

To swap the corners the first time, I applied B2 followed by the commutator J(b)-perm: R U2 F' R' F U' F' R F U' R' U'

Because commutators have to be in an even permutation, the commutator J-perm is actually a 3 cycle of pairs followed by an AUF.

A commutator requires that you insert something into a layer or space on the cube, then change it out, and then fix the rest of the cube by undoing the insertion, and then undoing the swap. I have to move a pair out of the U layer so it can be inserted in place of another. The move R does this. Doing a U2 makes the insertion easier.

You could insert the pair in place of another pair in the U layer with R', but you mess other parts of the U layer up. You must first move the URF corner out of the way with F'. This gives you F' R' F for the insertion.

To swap the inserted pair out with another, you can do U'. You then undo the insertion giving you F' R F. You now undo the swap giving you a U. You then undo the premoves giving you U2 R'. The U and U2 cancel to give you U' instead. At the end you AUF with a U.
To swap the corners back into the correct orientation, do a z' rotation and apply the mirror of the above J-perm: L' U2 F L F' U F L' F' U L U. You can then undo the roatation with a z and fix the back side with another B2.

At this point, you need to cycle the UF, RF, and RU edges with the commutator U M U' M' F' M U M' U' F.

The UR ewdge needs to got to UF. I You can place it there with a U, but you need your URF and RLF corner to stay where they were. You can put them back with a U' but first, hide the edge with an M. After the U', undo the M with an M'.

Next, do an F' to swap the RF edge with the UR edge so that the RF edge will end up in the RF position and the UF edge will end up in the RF position after undoing moves.

Now you undo the insertion with M U M' U' To finish the commutator, undo the F' with an F.

You know that the first edge you insert will be oriented correctly if you insert it that way. If you make the first insertion into a slot that is already oriented, (the UF edge) that slot will stay in the orientation it was in. The last edge must be correctly oriented because you cannot flip 1 edge.
Finally, you need to fix the centers with the commutator E' M E M'.

The E' moves the R center to the F centers position. The M places the U center in the F position taking the place of the R center.

When you undo the E' and M with E and M' the U centers goes back to where the R center was and the F center goes back to where the U center was.

The centers in back where cycled as well, but other than that, nothing else moved because you can only move the pieces where the part one and two of the commutator intersect.
The whole solution: B2 R U2 F' R' F U' F' R F U' R' U' z' L' U2 F L F' U F L' F' U L U z B2 U M U' M' F' M U M' U' F E' M E M'.
 
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