Lesson 1 · Blindfolded cubing

The letters, and your first trace

Turning a scrambled cube into a short string of letters. This is the half of blind solving that has nothing to do with your hands — and the half that transfers to the 4x4 unchanged.

Why this, before any algorithm

A blind solve is two separate skills wearing one costume. First you convert the cube into letters. Then you execute those letters. Beginners assume the algorithms are the hard part; they are not, and for you they will be the easy part, because you already have the hands.

The transfer is the whole reason we start on the 3x3. Wings on the 4x4 are solved with r2, which is M2 with the slice swapped — the wiki notes that most r2 targets are simply the M2 algorithms with M2 replaced by r2. The tracing you learn today does not change at all.

Tracing is worth learning first because it is the part that scales. On a 4x4 you will trace corners, then wings, then centres — three passes of the exact loop you are about to learn, on a puzzle with 24 wing stickers and 24 centre stickers instead of 24 edge stickers. Nothing about the loop changes. Only the count does.

One idea: the buffer

Pick one piece position and call it the buffer. For edges we use DF, because that is the buffer M2 uses, and because on the 4x4 the r2 buffer sits in the matching spot, DFr.

Every blind method boils down to one repeated operation: exchange whatever is sitting in the buffer with the piece at some target position. Old Pochmann does it with a T-perm, M2 does it with the move M2, r2 does it with r2. Different algorithms, same operation.

So a solve is a list of places to shoot the buffer to, in order. Each one is a target. The list is your memo.

Naming the 24 stickers

Targets are sticker positions, not pieces — orientation matters, so which sticker of the UB edge you mean is the whole point. Twenty-four edge stickers get twenty-four letters, under a scheme called Speffz that virtually everyone uses.

That rule generates all 24. Do not memorise a table — memorise the rule and read the rest off the cube. Here it is applied to the edges:

A sticker also has a plain name: its own face first, then the face its partner sticker is on. Speffz H is the L-face sticker of the BL edge, so its name is LB. Its partner across the piece is BL, which is Speffz R.

Two letters always name the same physical edge — A/Q, B/M, C/I, D/E, F/L, G/X, H/R, J/P, K/U, N/T, O/V, S/W. Knowing a sticker's partner instantly is most of what makes tracing fast later.

Ten questions is enough to meet the scheme, not enough to own it. Owning it takes a few minutes a day for a week or two — so that lives in its own tool: the Speffz trainer. It schedules the 48 stickers so you review each one just as you are about to forget it, times your recall rather than your typing, and only counts a letter as known once it comes out without thinking. Open it a few times a day; each sitting is short by design.

The tracing loop

Now the actual skill. It is four lines long.

1. Look at what is sitting in the buffer, DF.
2. That piece belongs somewhere. Find the sticker position where the sticker now facing D needs to end up. Write that letter down. It is your target.
3. Go and look at the position you just wrote down. Repeat from step 1 using the piece you find there.
4. Stop when the piece in the buffer is the DF edge itself — whichever way round it is facing.

The subtle bit is step 2: you match sticker to sticker, not piece to slot. If DF holds the UB edge with its white sticker facing down, the white sticker must end up on U, so the target is A — not Q.

It is worth seeing why the D-facing sticker is the one that names the target. A shot swaps the two slots sticker for sticker, in order: whatever sits at U lands on the target, and whatever sits at K lands on the target's partner. Naming H in the example above therefore sends orange to H and blue to R, in one move — both correct. Naming R instead would send the orange sticker onto the B face, and the piece would arrive flipped.

This is why "the blue one goes to R" is a true thing to notice and still the wrong answer. It describes the other half of the same swap. The target letter is always the destination of the sticker facing D — the buffer's own sticker position.

Worked example

Scramble:

Tracing the edges from DF gives these first four targets:

Cycle breaks

Step 5 above is the one case that trips people. The buffer has come home, but the cube is not finished — some edges were never touched by that cycle. So you break into a new one: pick an unsolved sticker, write its letter, and carry on tracing from there as normal.

In the example, the cycle closes after four targets, and the lowest still-unsolved sticker is F. The full memo comes out as:

H B J A  ·  F O G T W F

Notice the second group opens and closes on the same letter, F. That is normal: breaking into a cycle costs one extra target, because the first shot parks the buffer's own piece somewhere and the last shot fetches it back.

Ten targets, an even number — so this scramble has no parity. An odd count means parity, which needs one extra fix at the end. That is lesson 3's problem; for now just notice whether your count is odd or even.

When the buffer comes home flipped

Sooner or later DF will hold the DF edge the wrong way round — yellow facing front, green facing down. It is tempting to call the target K, since swapping the two stickers would fix it. But K is the buffer's own sticker, and a shot swaps two different slots. No single shot can flip a piece in place.

So treat it like any other homecoming: the piece belongs in the buffer, therefore the cycle is closed, therefore you break. Orientation does not enter into it.

Rare as an opening — the buffer is flipped on the very first look in under 4% of scrambles — but it turns up mid-solve in about two thirds of them, once cycle breaks start moving the buffer piece around.

The flip is not lost, and you never fix it directly. Breaking shoots the flipped edge out to some other slot; a few targets later it returns, correctly oriented, for free. Here is that happening, starting from a flipped buffer:

This is the general shape of blind solving: you never repair anything in place. Every piece is fixed by passing through the buffer.

Practise

Three rungs. Do them in order — each one removes a support the previous one gave you. Grab your cube and apply the scramble if you want, or just read the diagram.

Anything here that did not land — the D-face x2 convention, why orientation decides the target, when a break is allowed — ask me. That is what I am for, and a misunderstanding at this stage quietly ruins every solve later.