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While solving blindfold, I'm facing this rather icky situation where a cycle of edge or corner pieces end with an inverted piece. For example, if I start a cycle with B (White and Red) and it ends with M (Red and White). Do I need to remember this while solving or is there another way, because all I understand is that if I leave that piece as M, I'll find another inverted edge piece in the end to adjust this with.

For example, let us consider the following scramble : R' B' U' R' F' R2 D' R L' B' R' L D2 U2 L D2 U B2 R U F2 U R2 U F

I did this with green face as FRONT and white face as TOP. Now using usual nomenclature of edge pieces by Upper Case letters and corner pieces by Lower Case letter, I get a cycle as "BGLPOHUCDM". One may notice that B (WHITE-RED) and M (RED-WHITE) denote the same edge piece but in different order.

Hence, my question is : Is there any way to avoid such event or if not, how to tackle this with a little more style?

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  • $\begingroup$ Welcome to Puzzling.SE! Could you explain the problem in a little more detail for us Rubik's laymen? I'm not sure that I understand what you mean by "inverted" and I also can't quite picture the "cycle", although I think I understand what you mean. $\endgroup$ – Brandon_J Feb 25 at 14:25
  • $\begingroup$ @Brandon_J, thank you for responding. I have edited my question with more clarification. $\endgroup$ – Jyotirmoy Pramanik Feb 26 at 8:28
  • $\begingroup$ Makes much more sense now; thanks! Here, have an upvote! $\endgroup$ – Brandon_J Feb 26 at 14:41
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I believe I have solved it myself.

As it happens, I observed that one doesn't need to keep such alterations in mind as they tend to take care of themselves. This is true because moves on a Rubik's cube form a non-commutative group. Moreover, I believe (I am not completely sure but it's a conjecture) that the edge perms and corner perms on a Rubik's cube belong to different conjugacy classes and as a result this algorithm has a certain robustness.

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