I found this from an old thread. ; Using 2 extra moves to convert it to a solvable state seems reasonable. How do you generate skewb algs in sledges?I don’t believe so, but correct me if I’m wrong
Skewb has two sets of corners, fixed and hanging. Each of these sets forms a tetrad, and the permutation of each tetrad is independent of the other. However, the permutation within each tetrad is constant. There is only one way for the tetrads to be aligned, and when this happens, skewb CP is solved.
The cool thing is that skewb only requires at most two moves to align these tetrads. This is because two of the four corners of any given layer are solved because each tetrad has a fixed permutation relative to itself. To position the other two corners, only two moves are required. (Probabilities: 0 moves = 1/12, 1 move = 2/3, 2 moves = 1/4) This in turn finishes CP.
Remember, this doesn't mean the layers of corners are oriented; they're just permuted.
Once CP is solved, only sledgehammers and rotations are required for the rest of the puzzle. This can be proven by showing that CO and centres can be solved independently only using sledges and rotations.
Starting with centres (disregarding CO):
Where A = sledge
Arrows point to cases
U: A y2 A
H: A -> U
Z: A y' A y' A -> U
Ue-1/Ue-2: -> U
Oa: y' A -> U
Ob: A -> U
Za: y A -> U
Zb: y' A -> U
Xa: A -> Ob
Xb: A -> Oa
Swirl-a: A -> Zb
Swirl-b: A -> Za
Wat-a: A -> Oa
Wat-b: A -> Ob
Then corner orientation by sledge+rot. can be proven by knowing that A2 (rot.) A2 can be used to purely orient 2 diagonal corners of a layer. Since the CO mod3 rule applies within each tetrad, A2 (rot.) A2, aka "Pure Peanut", can be used repeatedly with rotations to orient all the corners.
The qualms I have with trying to use ksolve+ are that it counts rotations as moves. So, unless I wrote the how sledge affects skewb from 24 angles, I don't think I have the capacity to find God's number for sledges on skewb with solved CP.
It has been expressed to me that it's easy to prove that only five sledges/hedges are needed to solve first layer. Sarah's advanced variation includes solutions of max 5 sledges/hedges for each L5C+CLL case, so we can assert an upper bound of 10 sledges/hedges are required.
However, I found an approach to solving first layer that only requires four sledges/hedges, lowering the upper bound to 9 sledges/hedges.
Solving first layer with only four sledges/hedges:
1) Align tetrads (given)
This step only takes 2 moves max anyway, average 1.167 moves.
2) Solve the corners of one layer
If 4 corners are oriented: This step is solved.
If 3 corners are oriented: Use one sledge/hedge to change the orientation of the "bad" (unoriented) corner while keeping it bad, while making a good (oriented) corner from the layer bad. (These corners must be adjacent.) Now, two adjacent corners should be oriented and the remaining two adjacent corners should be unoriented. Follow the instructions for cases with two adjacent corners oriented.
If 2 adjacent corners are oriented: Use one sledge/hedge to orient the two adjacent corners without affecting the orientation of the two good corners.
If 2 diagonal corners are oriented: If you have Peanut CLL, you can solve that as normal. Otherwise, a general approach: Use one sledge/hedge to make one bad corner good while making one good corner (which is adjacent) bad. Then, you should have two good adjacent corners next to two bad adjacent corners. Now use one sledge/hedge to orient the two bad corners as above.
If 1 corner is oriented: In much the same way as the general approach if two diagonal corners are oriented, use one sledge/hedge to create a 2 adjacent case. Then use an additional sledge/hedge as before.
If 0 corners are oriented: If you have Pi CLL, use sledge/hedge once. Otherwise, orient pairs of adjacent corners using one sledge/hedge for each pair.
Maximum required sledges/hedges: 2.
3) Attach the centre
Use a pure U-perm to attach the centre, which requires either two sledges or two hedges.
Maximum required sledges/hedges: 2.
tl;dr Upper bound at 9.
Oh hey, that's awesome. I forgot I ever did that.I found this from an old thread.
You can use ksolve+ and generate algs involving sledges and rotations, but afaik each 'move' you define contributes 1 to the movecount. That is to say, "sledge y sledge" counts as 3 moves if "sledge" and "y" are defined moves.Using 2 extra moves to convert it to a solvable state seems reasonable. How do you generate skewb algs in sledges?
Most of them use weird optimal algs, but there are a few that use s/h but they have a lot of z/x rotations and they are 5-6 sledges long.Does it use moves outside of s/h? Any idea how to find that page?
Do a setup move and upermAre there any good algs for the 2 skewb cases where all the corners and 3 centers are solved, but you need to cycle the last 3 centers clockwise or counter clockwise accordingly? (Rubiks skewb Or WCA notation is finding) Its the 2 L3C cases (not the U-Perm) if that helps explain, wont let me upload a photo
Set up to Uperm yeah but more specifically:Are there any good algs for the 2 skewb cases where all the corners and 3 centers are solved, but you need to cycle the last 3 centers clockwise or counter clockwise accordingly? (Rubiks skewb Or WCA notation is finding) Its the 2 L3C cases (not the U-Perm) if that helps explain, wont let me upload a photo
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