Learning PLL properly
All twenty-one, recognised by headlights and blocks, and why this is the set where fluency pays back most.
By the end you will be able to
- Split the 21 cases into three families with one glance at the corners
- Tell the four G permutations apart, which is the hardest recognition in the set
- Take the cases in an order that keeps the recognition coherent
- Say why PLL repays drilling more than any other set
Twenty-one algorithms, and you use one of them in every solve you will ever do. No other set on the cube has that property: an F2L case might not appear, an OLL case turns up once in fifty-four solves, but the last layer always has to be permuted, and there are only twenty-one ways it can need doing.
That is why this is the set to make fluent rather than merely known. Taking a T permutation from two and a half seconds to one and a quarter is a second and a quarter off every single solve, for a fortnight of drilling and nothing new memorised.
How often each one comes up
Sixteen of the twenty-one turn up about once in eighteen solves — so if you solve twenty times a day, you meet each of them roughly every other day. Five are rarer: the Z and E permutations about once in thirty-six, and the H and the two N permutations once in seventy-two. A skip, where the layer is already permuted, is also one in seventy-two.
Recognition: corners first
Look at the two top-layer corners on each of the four faces and ask one question: do they show the same colour? A face where they do is showing headlights, and it means those two corners are already correct relative to each other.
- Four faces with headlights — the corners are done. One of the four edge-only cases: Ua, Ub, H or Z.
- Exactly one face with headlights — two adjacent corners need swapping. Twelve cases, including all four G permutations.
- No headlights anywhere — the two corners that need swapping are diagonally opposite. Five cases: E, V, Y, Na, Nb.
There is no other possibility. Two faces with headlights cannot happen, and neither can three, which makes this the cheapest question in the whole of CFOP: ask it of each face and twenty-one cases become four, twelve or five.
Then the blocks
A block is two neighbouring stickers of the same colour on one face — a corner and the edge beside it. A face showing all three the same is a solved bar: that face needs nothing doing. Counting blocks splits each family neatly.
| What you see | What it is |
|---|---|
| Four faces of headlights, one of them a solved bar | Ua or Ub — three edges cycling |
| Four faces of headlights, no bar | H or Z — the edges swap in pairs |
| One headlight face, a solved bar, blocks on the other three | Ja or Jb |
| One headlight face and two blocks | Aa, Ab or T |
| One headlight face and one block | Ra, Rb, or one of the four G permutations |
| One headlight face, a solved bar, and no blocks at all | F |
| No headlights and nothing else either | E — the corners swap diagonally, the edges are done |
| No headlights, a block on every face | Na or Nb |
| No headlights and two blocks | V or Y |
The four G permutations
These are the ones people leave until last and then find they cannot tell apart, which is a shame, because between them they turn up in about one solve in five. Learn them as a family of four, in one fortnight, using one consistent check.
Hold the case with the headlights at the back. There is now exactly one block anywhere on the cube, and it touches either the front-left corner or the front-right corner. Which corner, and which of that corner's two faces the block lies on, names the case.
| The block is | It is |
|---|---|
| On the front face, at the left — front-left corner with the front edge | Gd |
| On the left face, at the front — front-left corner with the left edge | Ga |
| On the front face, at the right — front-right corner with the front edge | Gb |
| On the right face, at the front — front-right corner with the right edge | Gc |
Uw turns that let the whole thing run without a regrip. There is a version with D turns instead if wide turns of the top layer feel wrong; it costs three more moves. The other three G permutations are close enough in shape that the fourth takes an afternoon once you have the first three.An order for the twenty-one
Recognition is the reason for this order rather than difficulty. Cases that are read the same way are learnt together, so each group closes a branch of the tree above.
| Order | Cases | Why here |
|---|---|---|
| — | Ua, Ub, H, Z, Aa, Ab | Already known from two-look |
| 1 | T, Ja, Jb | Short, common, and the blocks make them unmistakable |
| 2 | F, Ra, Rb | The rest of the adjacent-swap family bar the G permutations |
| 3 | Y, V | Diagonal swaps, two blocks each, told apart in one look |
| 4 | E, Na, Nb | The rest of the diagonal family; E may already be known |
| 5 | Ga, Gb, Gc, Gd | Last, together, with the check above |
U' included, because that turn is part of the solve whether or not the list you copied it from says so.Two-sided recognition
The goal is to name the case from the two faces you can see without moving the cube or your head. It is worth being honest about why this is harder than it sounds: the pattern of headlights and blocks on two faces does not always pick out a single case, and for some pairs of faces it leaves as many as eight candidates. What finishes the job is the colours — not that there is a block, but that the block is the blue one and the face beyond it is red.
So two-sided recognition is learnt case by case rather than as a rule, and it takes a few months. Start it early anyway. Every rotation you make to see the back of the cube costs you about a third of a second and, worse, arrives at a cube you have to re-read.
Drills that work on this set
- The time attack. All twenty-one in a row, timed, from a list. Under a minute is a reasonable first target; solvers with a fast last layer manage half that. It is the single most informative minute of practice in cubing, because the slow ones announce themselves.
- Four angles. Take one case and drill it from all four positions of the top layer, so that recognising it never depends on which face you happened to be looking at.
- Mirror pairs together. Ja against Jb, Ra against Rb, Ua against Ub, and — once you meet them — Ga against Gc and Gb against Gd. Drilling a case against the one it is confused with is worth more than drilling it alone.
- Recognition only. Set up a case, name it, do not solve it. Twenty cases a minute, and it is the drill that moves your times most in the week before a competition.