The Independence of OLL and PLL Parities

Being a cuber for eight years, I’ve come to know about the mathematics about the Rubik’s cube and the variants quite a lot, but there was one question I wanted to get straight: on a $4\times4$, if we have OLL parity, does that tell us anything about whether we’ll also get PLL parity? In other words, are those parities dependent to each other?

The usual answer is no. Each has a $50\%$ chance, so getting one doesn’t make the other any more or less likely. More precisely, this is true when we consider a uniformly random reduced state: the centres are solved in the correct colour scheme, and all the edges are paired. We’ll see what that means and why that qualification matters later.

But, just saying that they’re independent isn’t particularly satisfying. Both are called “parity,” both appear while solving the same cube, and both require us to interrupt our normal $3\times3$ solve with some extra moves. So, what are they actually measuring?

Pretending that a 4×4 is a 3×3

The usual way to solve a $4\times4$ is through reduction. We first build the six centres, then pair up the edge pieces, and finally solve the resulting cube as if it were a $3\times3$.

That last step is doing quite a bit of work. Each “edge” is actually made of two separate pieces, called wings. Once we pair them, we treat them as a single edge, sometimes called a dedge, or double edge. Similarly, four centre pieces become one larger centre. From then on, outer face turns keep these groups together, so we can use our familiar $3\times3$ algorithms.

Except, sometimes, we get something that a $3\times3$ can’t do:¹

• OLL parity: an odd number of paired edges appears flipped. The simplest example is a single flipped dedge.
• PLL parity: the permutation parities of the corners and paired edges don’t match. The simplest example is two dedges needing to swap while everything else is solved.

OLL stands for Orientation of the Last Layer, and PLL for Permutation of the Last Layer. Those are the stages where we usually notice these problems, though the conditions already exist when reduction is finished. Do note that we haven’t broken the cube. We’ve just reached a perfectly legal arrangement that our $3\times3$ moves can’t solve. The word “error” in parity error refers to that mismatch with the reduced puzzle, rather than a mistake we necessarily made.

What does even or odd mean here?

A permutation is a rearrangement of some objects. We can produce any permutation through a sequence of swaps of two objects, called transpositions. If the number of swaps is even, the permutation is even; if it’s odd, the permutation is odd.

The interesting part is that, although there can be many different sequences of swaps producing the same rearrangement, they all agree on whether that number is even or odd. We call this its parity, and write its sign as:

Doing two permutations in succession multiplies their signs. Two odd permutations give an even one, just as two odd numbers add to an even number.

For example, a quarter turn cycles four pieces around a face. Moving four objects around a cycle takes three swaps, so a four-cycle is odd, even though four itself is even! In general, a cycle of length $k$ has sign $(-1)^{k-1}$.

On a $3\times3$, an outer quarter turn cycles four corners and four edges. Both permutations are odd. A half turn performs each four-cycle twice, making both permutations even. Consequently, starting from a solved cube, the corner and edge permutation signs always match:

This is why we can’t swap just two edges on a $3\times3$: that would change the edge sign without changing the corner sign. Separately, legal $3\times3$ moves preserve an even total number of flipped edges, so a single flipped edge is impossible too.

These are two different restrictions, and as we will see reduction on a $4\times4$ doesn’t automatically satisfy either of them.

What are we actually moving?

Let’s look at the pieces individually. A $4\times4$ has:

• $8$ corners, just like a $3\times3$;
• $24$ wings, which we group into $12$ dedges;
• $24$ centres, four of each colour.

For the maths, imagine that every centre has its own tiny label, even when four of them have the same colour. Otherwise, we couldn’t distinguish their permutations. We’ll write $C$, $W$, and $X$ for the permutations of the corners, wings, and centres, respectively.

We also need to be precise about notation. Here, R means the outer right face, r means only the inner slice next to it, and Rw means both layers together. Likewise for U, u, and Uw. A prime reverses a turn, and 2 means a half turn.

The quarter turns have the following cycle structures:

A half turn is even in each of these three sets, since it performs the corresponding quarter turn twice.

Two useful things follow. First, the corners and the labelled centres always have the same permutation sign:

Second, only the inner-slice part of a move changes the wing sign. Outer face turns always move the wings through an even permutation. Starting from solved, if $n_{\text{inner}}$ counts all inner-slice quarter turns, including those inside wide turns, then:

For this count, a clockwise or anticlockwise quarter turn contributes one modulo two; a half turn contributes zero.

Notice that the centre equation doesn’t mean that solving the centre colours forces the corners to be in an even permutation. Two white centres can have exchanged places, and the centre still looks solved. Those tiny imaginary labels matter. Imagine the $3\times{3}$ with arrows on the faces, it is the same thing in this case too.

OLL parity is about the wings

Suppose we start with a solved $4\times4$ and exchange the two wings belonging to a single edge. Because a wing’s orientation is tied to its position, exchanging the two makes the whole dedge look flipped. At the level of the $24$ wings, we’ve performed exactly one swap: an odd permutation.

Now compare that with exchanging two whole dedges. If their wings are $A_1, A_2$ and $B_1, B_2$, the exchange involves two swaps:

That’s even! An odd permutation of the pairs can be an even permutation of the pieces inside those pairs.

More generally, let $\pi$ be the permutation of the $12$ dedges, and let $f$ be the number of flipped dedges under the usual $3\times3$ edge orientation convention. Permuting whole pairs contributes an even wing permutation, while each flipped pair contributes a swap within that pair. Thus:

So, OLL parity is precisely an odd wing permutation in the reduced state. This is also why outer face turns can’t fix it, since none of them changes $\operatorname{sgn}(W)$.

After solving the first two layers (F2L) as on a $3\times3$, the condition appears as one or three misoriented last-layer edges, rather than zero, two, or four. That’s when it tends to become obvious.

Is it already decided by the scramble?

Not by the scramble alone. We have to include the moves we make while solving the centres and pairing the edges:²

where both counts refer to inner-slice quarter turns.

Common edge-pairing sequences use a slice, some outer moves, and an undo of the slice. The two slice turns contribute an even number, preserving the wing sign. If all our edge pairing has that property, OLL parity is effectively decided once we finish the centres. But centre solving itself can change it, and a different reduction strategy can change it too.

This is actually exactly what makes deliberate parity avoidance possible. A cuber can trace the wing permutation and choose the parity of their centre solution accordingly, though I can never do that since it requires quite the heavy mental tracing.³

PLL parity is about the pairs

For PLL parity, we compare the dedge permutation with the corner permutation, just as we would compare edges and corners on a $3\times3$. Let’s define:

If $p=+1$, their signs match. If $p=-1$, we have PLL parity.

Every outer quarter turn changes both signs, so their product stays the same. No amount of normal F2L, OLL, or PLL can remove that mismatch.

But, remember the two-dedge exchange from before. Swapping two dedges changes $\operatorname{sgn}(\pi)$ while leaving $\operatorname{sgn}(W)$ unchanged. If the corners stay put, we’ve changed the PLL parity condition without changing the OLL parity condition.

Conversely, flipping one dedge exchanges its two wings, changing $\operatorname{sgn}(W)$, but it doesn’t move the dedge to a different edge position or move any corners. We’ve just changed OLL parity without changing PLL parity.

Can we actually change just one?

We can! There are legal move sequences for both operations.¹

A pure dedge flip, with the affected edge at the top-front, is:

1
r' U2 l F2 l' F2 r2 U2 r U2 r' U2 F2 r2 F2

It exchanges the two wings of that edge while leaving the other edges and corners alone and restoring the centre colours. Counting the inner-slice moves gives $9$ quarter turns, an odd number, as expected.

For a pure swap of the top-front and top-back dedges, we can use:

1
r2 U2 r2 Uw2 r2 u2

Remember that the final u2 is only the inner upper slice. Equivalently, we could finish with Uw2 U2. All the moves are half turns, so the wing permutation stays even, although the permutation of the dedges is odd.

There is a small trap here: not every OLL parity algorithm is a pure flip. Some also permute last-layer pieces, and some toggle PLL parity as well. The name only promises that they correct the orientation problem. So, “fixing OLL parity never changes PLL parity” would be too strong a statement. What matters for our argument is that we can choose an operation that changes only one condition.

From possible states to probabilities

Being able to reach all four combinations doesn’t, by itself, prove statistical independence. We also need to know how those states are being sampled.

Let’s take the model from the beginning: a uniformly random legal reduced state, with a fixed centre colour scheme. Define two bits:

The pure flip maps every state in class $(o,q)$ to one in $(1-o,q)$. The pure swap maps it to one in $(o,1-q)$. Both operations are reversible, so these maps are bijections: every state has exactly one partner in the other class.

Consequently, all four classes contain the same number of states. Under uniform sampling, they have the same probability:

Now we can actually justify the independence claim:

This equals $P(\text{PLL parity})$. The same calculation works the other way around, and conditioning on not having either parity also leaves the probability of the other at $1/2$.

Thus, the chance of at least one parity condition is:

So the familiar $75\%$ figure follows from the structure of the reduced states. It doesn’t mean that every four solves must contain exactly one of each case, of course.

What about bigger cubes?

The same distinction between individual wings and whole edges is useful on larger cubes, but we need to be careful about when we’re looking at the puzzle.

A wing orbit is a set of wings that can move into one another’s positions. Different distances from the middle of an edge give different orbits, and legal moves can’t transfer a wing between them. Each orbit contains $24$ wings. For an $N\times N$ cube with $N\geq3$, the number of wing orbits is:

On an odd cube, each edge also has a central piece, often called a midge. These pieces move around the cube, but they provide the $3\times3$ edge structure. Relative to the fixed centres, they obey the ordinary edge-orientation and edge/corner permutation restrictions. Once every wing is correctly paired with its midge, the resulting $3\times3$ stage has neither OLL nor PLL parity.⁴

The thing often called ”$5\times5$ parity” is a wing-pairing problem encountered before that reduction is complete. We can use algorithms related to the $4\times4$ OLL parity algorithms to fix it, but it isn’t a single fully assembled edge flipped during the final $3\times3$ stage.

On an even cube, there are no midges, so both OLL and PLL parity remain possible after reduction. More wing orbits give us more wing-parity information to deal with during pairing. However, once all the wings form complete edges, every orbit must agree on each whole edge’s orientation, and hence on its total flip parity. A fully reduced $6\times6$ has one whole-edge OLL parity condition, not two independent ones waiting in its last layer.

Overall, the answer to the original question is quite neat. OLL parity tells us about the wings, PLL parity tells us about the edge pairs relative to the corners, and neither forces the other. Under the usual model, seeing that flipped edge gives us exactly zero extra information about whether a pair swap is coming next.