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pattern

I know one that there was a similar one called "glitch".

If possible, could you also provide notations/a video tutorial for making it as well.

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2 Answers 2

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This (and many many other permutation puzzles) is easy to solve once you understand how to use commutators and conjugates. In this case, we need only 1 setup move and a 8-move commutator (2 of which are slice turns), which amount to 10 moves under slice-turn metric and 12 under face-turn metric:

F ( U' M2 U ) R ( U' M2 U ) R' F'. (The commutator is the middle 4 blocks.)

Of course, to set it up, just undo the sequence. This is superior to rhkoulen's solution under any of the usual metrics.

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  • $\begingroup$ So I used cube explorer to generate the most simple version of it but thanks youtu.be/svh5ceX20gw $\endgroup$
    – Scriptonix
    Commented Nov 10, 2021 at 16:18
  • $\begingroup$ Okay, so the solution you found via the cube explorer takes 13 face-turns or 14 quarter-turns, whereas mine takes 12 face-turns or 16 quarter-turns. So if you are looking for minimum quarter-turns then yes it is probable that yours is optimal. On the other hand, solutions with minimal quarter-turns tend to be inscrutable, and I don't usually understand them. If you want to be able to solve arbitrary patterns relatively efficiently (even if not optimally), you should learn from the linked post! By the way, you can accept if you are satisfied with my answer. =) $\endgroup$
    – user21820
    Commented Nov 10, 2021 at 16:28
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Used a solver to brute force this, but user21820 has since come up with a better solution...

D' F2 L2 U' L2 D F2 U' R' B U B' U R U2 R U

If you care, I found the "Glitch" as well.
It can be achieved with:

F' L' B' R' U' R B L F U

There are also a few YouTube tutorials for this if you can't read notation.

Note: I found this one on Ruwix. They have two nice pages of patterns (1) (2), and this was on the second page.

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    $\begingroup$ See my answer for a better solution! Also, I'm not sure what you're asking for but it is impossible to just swap two edges due to the parity (see the linked post for an explanation). $\endgroup$
    – user21820
    Commented Nov 9, 2021 at 14:29
  • $\begingroup$ Thanks! I'm not really that into Rubik's cubes I was just dropping a quick solution. $\endgroup$
    – rhkoulen
    Commented Nov 9, 2021 at 19:26
  • $\begingroup$ You're welcome! =) $\endgroup$
    – user21820
    Commented Nov 9, 2021 at 19:41

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