Is there a Secret PLL?!

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@nanamacapagal8342 2 жыл бұрын

"Are there two E perm algs?" There are technically two ways to think about this. There are either 22 or, 72 PLLs in total. If I count 22, I'm counting the standard 21 plus the PLL skip (which I will be calling AUF from now on) . Simple enough. Where the interesting thing starts to happen is how you explain that there are 2 distinct E perms. If we're speaking like that, then there should actually be 72 PLLs. Here's why: Forget everything you know about the PLLs and just focus on how the last layer could possibly be put together. We can place the 4 corners wherever the hell we want, then place 2 of the edges wherever the hell we want. But the last two edges are forced so we don't wind up with an impossible PLL. Do the maths and you get 4! (for the corners) * 4! (for the edges) / 2 (to fix parity) = 288. These include the two distinct E perms you talked about as well as some other wacky shit like a random A perm + Z perm combo. Except we counted way too many. If you'll notice, if you actually do an Aa perm, then a Z perm, you wind up with a T perm (or F perm depending on which angle you do the Z perm part). How do we fix this. Remember when you said that there is a specific way to hold each PLL correctly? It turns out that the exact definition of "holding something correctly" is completely arbitrary. I can just as easily say "Let the red-green-yellow corner always be solved" and that would be a valid way to hold the PLLs too. So since the red-green-yellow corner is always solved we instead have: 3! (corners left to place) * 4! (edges left to place) / 2 (to fix parity) = 72. That's where you made your mistake: you declared two distinct cases for E because you didn't clearly define how to hold the case correctly. The fact that the probabilities of the PLLs are marked as a multiple of 1/72 (1/72 for H, N, and AUF; 1/36 for E and Z; and 1/18 for the rest) is not a coincidence. The 72 PLLs I described are all equally likely to show up. It's just that from our perspective some of them don't look that distinct. Four of these 72 all belong to the T perm: one where the headlights are solved and three where they aren't. 4 of these belong to the V perm: one where the block is solved and 3 where it isn't. It turns out that E perm only takes up 2 of these 72 instead of the standard 4 for most of the other PLLs: these two being the secret PLLs you described in the video. In this way, N perms are interesting. They only take up 1 of these 72 PLLs each: that's why they're so rare. You claimed there were only two correct ways to hold an N perm, but in fact there are four correct ways to hold it, one from each side. But how do I know this. How do I know that the N perm works from all angles but the T perm and Y perm only work from one? Simple. I do the same standard alg from all 4 different sides and check if it solves the layer. In the cases of H, AUF, and both N perms, it solves the layer regardless of where you hold it. In the case of E and Z, it only works from 2 sides. And for all the rest, they only work from one angle. So how did we get from 72 to 22? Simple: in the same way your 2 distinct E perms are both called singular E perm just because one can turn into the other by U moves and y rotations alone, the 4 distinct "T perms" all collapse into one singular T perm by means of U moves and y rotations. In fact, if any group of PLLs can turn into each other by means of U moves and rotations alone they are labelled the same. So under our "one corner solved" definition of PLL, two of these would look like a T perm and the other two would look like an A perm and a Z perm smushed together. So that's how we wound up with 22 PLLs. F, J, T, R, G, A, Y, V, and U all take 4 of the 72 cases each, Z and E take only 2 each (of which you only described the E cases), and H, N and AUF all take 1 each. Most people leave out AUF though so that's how we wound up with a more familiar 21.