# Skewb theory challenge



## Stefan (Dec 16, 2013)

Some Meffert's puzzles, including Skewb, have corrugated stickers. A fine line texture going one of two directions. You can somewhat see it from the pictures here:
http://www.cs.brandeis.edu/~storer/JimPuzzles/ZPAGES/zzzSkewb.html

Imagine a sticker on a corner. You can distinguish the two possible texture orientations by feel, so a corner sticker orientation can be marked and identified with this. If you call the two markings 0 and 1 and read clockwise around a corner, you can mark the corner one of four ways: 000, 001, 011 or 111 (because you don't know where to start reading, 010 and 100 are equivalent to 001, and 110 and 101 are equivalent to 011).

I believe with certain marking patterns, I can solve all corners just by feel. Can you? Or can you solve even more than that?


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## Sa967St (Dec 17, 2013)

Spoiler



Assuming only the corners are marked, you certainly can't solve beyond the corners, since you can end up with three 2-cycles of centers. 

Short answer to the problem: Yes, there is such a marking system so that we can solve all the corners.

Solution:

Starting with a more specific problem: How far can you get if you mark all corners as 100 where 1 represents white/yellow and 0 represents other colours?

We can reduce it to having all corners oriented with respect to U and D with this marking system. Once you have all the corners oriented on the D layer, you can reduce the corners on the U layer to have at least 2 of the 4 corners oriented.
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Case 0: All 4 corners on U are oriented. You're done.

Case 1: 3 of the 4 corners on U are oriented. Put one of the U corners beside the misoriented U corner beneath the misoriented U corner. There's a 50% probability that you can orient all the U corners, and a 50% probability you'll end up with Case 1 again. You can argue that this is never guaranteed to terminate, but c'mon, if you do it enough times you'll surely get it.

Case 2: 2 of the 4 corners on U are oriented. You can reduce it to be the "headlights" pattern as opposed to the "chameleon" pattern. Put down both of the misoriented corners to the D layer with one move each, then you can orient all the U corners.
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So it is indeed possible to reduce it to all corners oriented with respect to U and D. These states for the corners can be broken down into three cases.
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1. All corners solved,
2. Two diagonal corners swapped on D and two diagonal corners swapped on U.
3. Two U corners on D and two D corners on U.
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We can adjust the marking system so that all corners are marked 100 or 110, where white corners are 100 (the white sticker on each corner is marked 1), and yellow corners are marked 110 (the yellow sticker on each corner is marked 0). Then we can use the same approach to orient all the corners and have the white and yellow corners in their correct layer. This would be one of two states, 1. and 2. listed above.

We can adjust the marking system yet again so that we can avoid having a state where all corners are oriented and in their correct layers, but are not permuted correctly. This can be done by marking the white-green-orange corner as 000 and white-orange-blue corners each as 111.

Here's how that works:
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Step 1) Complete the white layer so that all of the corners are permuted with respect to each other, and the white-blue-red and white-red-green corners are oriented correctly.

Step 2) Orient as many of the yellow corners as possible.

Case 0: You can orient all of them. You're done -- all the corners are solved.

Case 1: You can only orient three of the four corners.

Case 2: You can only orient two of the four corners.

Step 3) If you had Case 1, one of the two possible D corners are misoriented. Take one of the corners out of the D layer and put it back in twisted clockwise from how it was before. Go back to Step 2. 

If you had Case 2, both of the two possible D corners are misoriented. Take both of the corners out of the D layer and put them both back in twisted clockwise from how they were before. Go back to Step 2.
_

You'll eventually terminate after the white-green-orange and white-orange-blue corners are both oriented correctly, giving a state where all corners are solved.


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## bobthegiraffemonkey (Dec 17, 2013)

Does this basically mean I'll have to resticker my skewb to make it comp legal? The meffert's colour scheme is weird anyway.


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## Stefan (Dec 18, 2013)

Nice, Sarah, that's a simpler scheme than I had in mind. I'm just surprised by your half-random orient-and-repeat-if-not-done strategy. Why not use an alg orienting two certain corners as desired? Like, in your final scheme, orient each misoriented yellow corner along with the 000 or 111 corner.

Mine was to mark URF with 000 and DLB with 111 and the three other of each tetrad (*) with a combination of 001 and 011 to determine and solve tetrad orientation and individual corner orientation, and URF and DLB will get oriented correctly automatically.

(*) group of four corners not moving relative to each other, Skewb has two tetrads



Sa967St said:


> Assuming only the corners are marked, you certainly can't solve beyond the corners, since you can end up with *three 2-cycles* of centers.



You meant *two 3-cycles*, right?

And yes, if you don't mark centers, of course you can't solve them the least bit. But what if you do mark them?



Spoiler: Thoughts about centers



After solving corners, use them as reference. You can't orient centers in place by 90 degrees, so you can mark each center as 0 or 1 (for each center, you can for example imagine permuting centers so that this center gets on U, and it will always be the same orientation there, which you call either 0 or 1). We can mark the six centers different ways:

000000 doesn't tell you anything.
000001 identifies one center, which you can then solve. There are 5!/2=60 remaining possible cases (how many after symmetry reduction?).
000011 identifies a group of two centers and a group of four centers, so (2!*4!)/2=24 cases (how many after symmetry reduction? Might depend on whether the group of two is opposite or adjacent).
000111 identifies two groups of three centers each, so (3!^2)/2=18 cases (how many after symmetry reduction? Again might depend how you choose the groups).
001111, 011111 and 111111 are equivalent to former ones.


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## Stefan (Dec 18, 2013)

This could be a magic trick, for example:
1. Have someone scramble your Skewb.
2. Solve behind your back as far as possible, claiming something like reading their mind for how they scrambled.
3. Present the Skewb.
4. If not solved, quickly identify the case, say "Darn, made a mistake", and immediately put it behind your back again and solve it. (Once I did a 3x3 BLD demo, made a mistake and thus presented an unsolved cube, but happened to see right away how to fix it, and quickly did so above my head. The audience was very impressed and the moderator asked whether that was on purpose and that I should do that on purpose )


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## Dapianokid (Dec 18, 2013)

I wonder if people with seriously messed up stickers are ever pulling one over on us by solving some parts by feel 
Stefan, this is genius. Of course YOU'RE the one getting us all worked up for SkewbBLD


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## Stefan (Dec 18, 2013)

Dapianokid said:


> Of course YOU'RE the one getting us all worked up for SkewbBLD



Well, Skewb was the second twisty puzzle I solved blindfolded, about 10 years ago (had done 2x2 before).


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## Sa967St (Dec 19, 2013)

Yes, I definitely meant two 3-cycles, or one 2-cycle and one 4-cycle. I wrote three 2-cycles without checking if one even exists, oops. 

The first half of what I wrote about 100 isn't necessary, but that was the most fun part to figure out!  There's probably a much more elegant proof of why the final scheme I came up with works.

(I purposely didn't quote your post because then I would be able to see what you wrote in the spoiler about centers. I want to figure that out. )


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