2 wings and 2 t-centers unsolved (quarter slice turn to fix parity)ģ.The Professor's Cube was invented by Udo Krell in 1981. 2 corners, 2 midges, 2 x-centers and 2 t-centers unsolved (quarter face turn to fix parity)Ģ. So you can end up with one of the following four scenarios in a 5x5x5 supercube BLD solve if you use ONLY 3-cycle commutators of each piece type:ġ.

But this is what I think it means: 4 distinct scenarios in which a 5x5x5 supercube can't be solved using only 3-cycle commutators of each piece type.Įach piece type on a solved cube has even permutation parity.Ī quarter face turn causes the following to have odd parity: Corners, midges, x-centers and t-centers.Ī quarter 'non-middle slice' (don't know how else to name it) turn causes the following to have odd parity: Wings and t-centers. Like I said, I don't exactly know what you mean by '4 distinct scenarios considering the parities of all piece types'.

A quarter turn will only toggle parity of a piece that is either in the middle of a face (along that slice) or on an edge, which is why it's the +centers and midges but not the xcenters or obliques that get parity. If you don't fix centers then you could consider an M move to have created parity on centers and midges, and it gets more complicated.) An R turn toggles parity of corners and midges, while a 2R turn toggles parity of wings and +centers. (Assume the centers are fixed, so M is impossible. (So, if you are solving a supercube 5x5, you actually never get the wing-transpose parity! Neat, huh? And a supercube 4x4 doesn't have +centers, so you can still get (both kinds of) parity after the centers are solved.) Normally the first set is only considered to be a parity in BLD, though.Īnd similarly #1 and #2 are equivalent and so on.Īnd about that 'simple method' - just consider the possible slice moves For instance, for the supercube 5x5, you have R and 2R. What I mean by that is that if you fix parity problem #1, parity problem #2 is automatically solved, and vice versa. The (supercube) 5x5x5 has '4' parities - that is, there are 4 types of pieces which can get a parity:īut as it turns out #1 and #2 are equivalent, and #3 and #4 are equivalent. There really isn't any point in learning algs for these cases because they occur so rarely that you don't really get to practice them at all and you'll probably take longer to recall them than to do 2 algs.īesides, for cases like D and G, you probably would've recognized them while pairing tredges and fixed them while it was still easy. If you get cases C, D, E, F or G, consider yourself unlucky and do the alg for either A or B to reduce the case to A or B. So you only need to learn one alg to solve B if you're doing AvG tredges. So actually, the only cases that ever appear are A and B (and of course H). (Someone double check this for me please!) With AvG, it's a little different: Cases C, D, E, F and G can only occur if they were already like that right from the start, i.e. With the tredge-pairing method, the probability of ending up with each case is obvious, you just have to count the number of cases in each group: I suck at mirroring on the fly (be it L/R or F/B), so this is definitely good for me. So in other words, excluding the solved case, there are only 7 algorithms needed for the last two tredges, and there isn't even a need to mirror any of them.

And to flip one of the tredges, you need an average of exactly 2 moves, you can see how I arrived at this number later on. All cases within the same group can be transformed into the asterisked case either by doing a U2, or by flipping one of the tredges.

5x5 rubix cube flip middle