Intuitive EOLR with two misoriented edges

A comprehensive guide by Pedro Ilídio.

1. Introduction

With a naturally lower move count, Roux has been considered a prominent alternative to the mainstream CFOP method for speedsolving. But for this potential to be fully explored and optimal times to be achieved with Roux, one of the essentialities is to employ optimizations of the last six edges (LSE) step, such as EOLR, increasing the move efficiency as much as possible but avoiding an expressive recognition burden.

However, despite Roux being advertised as an intuitive solving method, the learning of EOLR still usually involves memorizing and recognizing a bunch of specific cases and sequences of moves, even if parts of these sequences are commonly regarded as intuitive.

In this document, I present some ideas to improve EOLR intuitiveness that I have been experimenting with for a few months. Specifically, a general algorithm is proposed to obtain a “good arrow” from any of the LSE states with only two misoriented edges (2mo). Since a 2mo or 4mo state appears in 15 out of 16 times (93.75% probability), having intuitive solutions to these two states is of great relevance to one’s performance. Although finding a good arrow from 4mo in general is already pretty intuitive, new ideas for tackling 2mo are appreciated.

I do not expect the present procedure to be a replacement for the current recognition methods of experienced solvers, but I do hope to:

  1. Facilitate EOLR learning for newer Roux users, that still have limited to none experience with EOLR;
  2. Promote development of new LSE recognition techniques, proposing a framework of thought to be expanded by the community. Hopefully a humble step towards fully intuitive one-look LSE.

No EOLR knowledge is needed, but we require the reader to be familiar with Rubik's cube notation and the Roux method.

A throughout introduction to basic concepts of edge orientation in Roux’s last six edges (LSE) step is firstly presented in Section 2, aimed at describing the process of thought that originated these ideas and building the conceptual foundation for the final proposed algorithm, summarized at the end (Section TL;DR). An introduction to EOLR is then provided (Section 3), contextualizing EOLR in the framework described in the first setions followed by several case examples with links to 3D visualizations (Example solves appendix) and a final discussion of current limitations of the technique and open paths for future thoughts (Section 5).

Cube illustrations were rendered with PuzzleGen (rotations: {x:63, y:40, z:39}, strokeWidth: 0.04).

2. Edge orientation principles

Consider the much more simple setting of six light bulbs that can be either turned on or turned off. They start all turned off, and your goal is to light them all up. However, you are required to do so by an arbitrary number of steps in which, for each step, you select four lamps and toggle their switch, so that those which are turned on become turned off and vice-versa. The solution one would probably come up with is pretty straightforward: just turn 4 of the lights on, then select 3 lights on and 1 off to toggle, after which you end up with 4 lights off that you can just turn on in one last step (Figure 1).

Figure 1: Simple light bulb analogy to LSE.

The LSE-4a step of the Roux method is not that different from that simple analogy. In this step, each of the six edges can be either oriented or misoriented.

Definition: We say an edge is oriented if its U or D sticker (here represented as white or yellow, as many cubers choose it to be) faces up or down when the U or the D center is placed up or down (Figure 2).

Figure 2: Examples of LSE states with labels indicating the oriented (o) and misoriented (mo) edges.

Whenever a U turn is made around an F or B center (green or blue in the diagrams), the four edge pieces surrounding it switch orientation, so that oriented edges become misoriented and vice-versa (Figure 3). Since there are 6 edges remaining and we can only toggle orientation of four edges at a time, there can only be 2, 4 or 6 misoriented edges in total.

Figure 3: Illustration of edge orientation toggling. When a U or U’ move is made with an F or B center on top, the four edges around it flip their orientatons, so that the oriented edges become misoriented and the misoriented become oriented.

So the way we solve the orientation of the LSE necessarily involves a last stage where 4 remaining misoriented edges are oriented by a U or U’ move around a B or F center (similarly to the light bulbs analogy).

Notice that two consecutive turns (a U2 move) cancel each other out, having no effect in edge orientation.

The possible adjustments of the edges’ positions that do not change their orientation: 1. U or U’ can be arbitrarily performed if a white or yellow center is on top. 2. One can arbitrarily raise a red/orange center and do U2.

Since we usually recognize cases with white or yellow on top, it is useful to, in one of these states, consider beforehand the set of pieces that would be around the F or B centers if F or B is placed in the top layer. For the F center, this group of edges corresponds to the edges in the positions UF, UL, UR and DF. I will thereafter refer to this set of pieces as the F-cycle (Figure 4).

Analogously, I will call by B-cycle the edges at UB, UL, UR and DB collectively. Again, these cycles correspond to the pieces that will end up on top around the F center when an M’ move is performed (in the case of the F-cycle), or around the B center when an M move is executed (for the B-cycle). When the respective center (F or B) is on top, performing a U or U’ move will switch the orientation of the 4 edges of the corresponding cycle, besides shifting their position as shown in Figure 4.

Definition: We collectively call by F-cycle the edge positions UF, UR, DF and UL, and by B-cycle the edge positions UB, UL, DB and UR.

Figure 4: F-cycle representation. If M'UM is performed, "turning the F-cycle clockwise", the four highlighted edges will toggle orientation and be permuted as indicated by the arrows.

Therefore, the cycles correspond, in the simpler light bulb analogy, to the two possible sets of 4 lights that we can toggle together, and the last step of edge orientation also involves selecting 4 lamps off (misoriented edges) to turn on, corresponding to the usually called “arrow” configuration, where all edges of a cycle and only them are misoriented.

A difference here is that edges can be repositioned with arbitrary U moves around a white/yellow (U/D) center or U2 around  an F/B center, without changing their orientation.

Definition: I refer to performing M’ followed by U or U’ and then M or M’ as turning the F-cycle, and to making an M move followed by U or U’ and then by M or M’ as turning the B-cycle.

These optional turning directions will later be restricted by EOLR.

So the procedure to orient all LSE using only M and U moves (footprint: usual EOLR algorithms may involve R or r moves in the 4mo state, not considered here.) is strikingly similar to Figure 1. Starting from a state where all 6 edges are misoriented (what I will call the 6mo state), follow Procedure 1.

Procedure 1: orienting all the last six edges from the 6mo state.


  1. Switch 4mo edges.
    1. Turn the F- or B-cycle in any way. 2 misoriented edges remain (I will call it the 2mo state).
  2. Switch 3o + 1mo edges.
    1. Adjust the U face so to set the F-cycle or B-cycle to be composed of 3 oriented and 1 misoriented edge.
    2. Turn the cycle set in the last substep, switching orientation of the edges in the cycle. 4 misoriented edges remain (4mo state).
  3. Switch 4mo edges.
    1. Set the 4 remaining misoriented edges in a single cycle, by only turning the U face around the white-yellow (UB) centers or making U2 around a red-orange (FB) center, so that the edge orientation is kept.
    2. Turn the cycle composed of the 4 misoriented edges, orienting them.

Notice how the cycle-setting followed by cycle-turning substeps closely resembles the light bulb selection followed by light toggling in the previous simple analogy.

If there are only 2 or 4 misoriented edges in your cube, not 6, you may just skip to step 2 or step 3 above, respectively, but the order of the steps is definite. One can demonstrate that the states in the shortest route to edge orientation (or all lights on) have the imperative order 6mo ⟶ 2mo ⟶ 4mo ⟶ 0mo by enumerating the few possible state transitions (Figure 5).

Figure 5: State transition graph of edge orientation states for LSE. “mo” stands for misoriented edge, and “o” means oriented edge. Arrow labels indicate the composition of the selected group of edges to flip. For example, 3o + 1mo means to select 3 oriented and 1 misoriented edge and toggle their orientation. The shortest route to the goal 0mo state is highlighted in purple. These are the only relevant transitions, corresponding to the steps described in Procedure 1.

2.1. Arrow case lookahead from 2mo

A relevant property to be noticed is that, after step 2.a, the two remaining edges outside the set up cycle will necessarily be one oriented and one misoriented. Pay attention for a moment to two specific pieces: the single misoriented edge in the cycle you selected (which will become oriented after turning the cycle) and the remaining oriented edge outside the selected cycle. These will be the 2 remaining oriented edges outside the last cycle, formed in step 3.a. Notice that after step 3.a they will necessarily be separated by a blue or green (F/B) center. By recognizing these two edges early on and predicting their final position after 3.a, one can lookahead all the steps following 2.a.

To facilitate the explanation, I will refer to these edges in the following sections as the in-cycle misoriented (ICM) edge and the out-cycle oriented (OCO) edge. Remember that ICM and OCO in the 2mo state will become the only oriented edges in the 4mo (arrow) state.

Definition: The only misoriented edge in the cycle selected in step 2.a of procedure 1 will be here called in-cycle misoriented (ICM). The only oriented edge outside this cycle will be called out-cycle oriented (OCO).

Another thought I find useful in this process of looking ahead in step 2 is that ICM and OCO need to end up one in the U face and the other in the D face, i.e. one must ensure they are not left in the same U or D face.

  1. ICM and OCO will end up one in the U face and the other in the D face.
  2. ICM and OCO will also end up both in the same F or B face, separated by an F/B center.

3. EOLR principles

Although the Algorithm 1 to orient the LSE is inescapable, there is some degree of freedom in its steps that can be exploited to skip or simplify the Roux’s 4b step. This freedom manifests itself in two ways:

  1. There are usually multiple ways of performing the 2.a and 3.a and steps, i.e. one can set several cycles that satisfy the edge composition condition of the steps.
  2. In the cycle turning steps, are arbitrary both
    1. the direction (U or U’) in which we choose to turn a cycle after raising its corresponding center to the U face and
    2. the final choice of an M or M’ move to bring back a white or yellow center to the top.

The EOLR technique, first proposed by James Straugham (Athefre) and further developed by Iuri Grangeiro, Kian Mansour and Louis de Mendonça (see the original post and Speedsolving wiki ), consists of utilizing this freedom in order to place, during the edge orientation step (Roux’s 4a), the UR-UL or UF-UB pairs of edges in the DF and DB positions, where they can easily be inserted to their correct spots in the U layer, essentially combining Roux’s 4a and 4b in a single step, i.e. doing 4b while solving 4a.

Definition: Let’s generically call the original UR-UL or UF-UB edge pairs by matching edge pairs from here on.

Thus, the goal of EOLR can be stated as placing a matching edge pair in the DF-DB positions while solving LSE orientation.

Note: No distinction is made between UR-UL or UF-UB pairs, so that using the UFUB approach in Roux's 4b is naturally covered.

If we reverse engineer the solved state, with all edges oriented and a matching pair at DF-DB, we arrive at only 5 possible edge pair configurations in the 4mo state that allow for easy insertion at DF-DB while still solving orientation. And remember that, since 4mo with a cycle with only misoriented edges (an "arrow") is the mandatory last step for edge orientation, reaching one of these “good arrows” is mandatory to solve EOLR.

Figure 6 The 5 “good arrows” possible. The symmetric equivalents obtained by flipping B and F are not represented, but are just a y2 move apart from the examples. The original UR and UL (red-yellow and orange-yellow in the figures) are highlighted. In the 2R and 2L cases, the UL edge is at the DB position. Among the original UR and UL, which edge is UR and which is UL is not important in the figure, their positions could be swapped without compromising the diagram’s meaning.

Good arrow 1L - Set up: U M2 U2 M U' M - Solution: M' U  M' (3D view)

Good arrow 1R - Set up: U' M' U2 M2 U' M - Solution: M' U'  M' (3D view)

Good arrow 2L - Set up: U M U2 M2 U' M - Solution: M' U'  M (3D view)

Good arrow 2R - Set up: M2 U M' U M U2 M2 U2 - Solution: M' U  M (3D view)

Good arrow 3 - Set up: M U' M2 U2 M' - Solution: M U2 M2 (3D view)

Reverse engineering one step further, from the 2mo state (step 2 of Procedure 1), the goal thus becomes achieving not only a 4mo state but a good 4mo, i.e. a good arrow. And there lies the main objective here: searching for an intuitive procedure to solve EOLR from 2mo, with no need for memorization of any specific cases and their corresponding move sequences.

4. EOLR with 2 misoriented edges

From the 5 good arrow cases (Figure 6) we will concentrate only on the four first. In all the first four, notice how we always have one of the two oriented edges pairing with a misoriented one. The two oriented edges can be predicted from the 2mo state as described in section Arrow case lookahead.

So, besides the in-cycle misoriented (ICM) and the out-cycle oriented (OCO) edges described in the section, for EOLR, we also want to look for a third edge that matches up with one of them, so that this third edge and the ICM or this third edge and the OCO are either the original UL-UR or the original UF-UB edges.

At this point, the following statement can be made:

Conjecture: if there is a third edge pairing up with either ICM or OCO and if it is adjacent to the ICM in the cycle set up in the 2a step of Procedure 1, there necessarily is a rotation of this cycle that leads to a good arrow.

Given its importance, I will call this third edge, being the match-up with ICM or OCO, misoriented, in-cycle and adjacent to the ICM, by MICA. By “adjacent” in this context, I mean that, if the cycle is raised to the U face, this edge will not form a straight line with the ICM and the raised center, forming an “L-shaped”, 90º angle instead. This is equivalent to saying that the MICA and the ICM must be connected by one of the arrows illustrated in Figure 4.

Definition: We call adjacent positions in the F-cycle the position pairs UL-UF, UF-UR, UR-DF and DF-UL. The adjacent positions in the B-cycle are UR-UB, UB-UL, UL-DB and DB-UR.

Definition: In the 2mo state, an edge that matches either the ICM or the OCO, is misoriented, in-cycle, and adjacent to the ICM is called MICA.

Definition: An F/B cycle in the 2mo state that satisfies the already stated conditions of:

  1. Being composed of 3 oriented edges and only one misoriented edge (the ICM);
  2. Having a MICA.

is therefore called a good cycle (see Figure 7).

Figure 7: Three examples of good cycles are illustrated, with edge colors indicated besides labels and connectors indicating the MICA and its matching edge, both of which will end up in the DF and DB positions after EOLR is completed. The solutions to these and other examples are fully described in the Example solves appendix.

Set up:

U M U M2 U2 M U2 M' U M' U (Example case 5, first above)

M U' M U' M U M2 U M (Example case 4, second above)

U' M U' M' U' M U M U' M2 (Example case 6, third case above)

Finally, the rotation that turns this good cycle into a good arrow can be performed as described by Procedure 2, which substitutes step 2 of Procedure 1, specializing it.

Procedure 2: converting the 2mo state to a good arrow.


  1. Set up a good cycle;
  2. Raise the selected good cycle by performing M' if it is the F-cycle or M if it is the B-cycle;
  3. Execute a single U or U’ move in way that places either ICM or MICA (whichever is outside the M slice) diagonally opposite to the OCO (i.e. setting ICM and OCO, or MICA and OCO, to be separated by the cube's core);
  4. Do an M or M’ move in a way that keeps ICM and OCO in different vertical layers (one in the U layer, the other in the D layer).

After solving the good arrow, the MICA and its matching edge should arive at DF-DB.

Several examples of application of this procedure are shown in the Example solves appendix, with the intention of clarifying any confusion that may arise.

5. Final remarks and future work

Although less frequent, especially if employing UFUB, there are cases where no good cycle can be formed but a good arrow can still be obtained with the simple rotation of a cycle. For instance, if the ICM matches the out-cycle misoriented edge there will be no MICA-ICM pair (see the following example). However, they can still be paired up in a good arrow by performing M U M in the example below.

Example: 2mo EOLR unexplained case. - Set up: M U M U M' U M U' M ' U M U' - Solve: M U M (3D view)

An intuitive explanation is still lacking for those cases, so expanding or generalizing the proposed procedure should be possible.

Furthermore, the last "good arrow" case in Figure 6 is not sought by the proposed procedure, further suggesting eventual extensions.

Additionally, this work is written mainly from a "U/D center on top" (or “oriented centers”) perspective, so considering what is commonly called "misoriented centers" poses a new territory to be explored under the proposed optics. With F/B on top, turning the F/B-cycles, which now surround U and D centers, does not toggle their orientation, and the new goal of 4a seems to be generating 4 misoriented edges and getting them to the top in order to orient them with an U move. Employing moves that do not change orientation (U2 and cycle rotations) seems especially useful to bring edges to the top.

Regarding recognition time, learning each EOLR case separately should naturally still be faster. However, the proposed framework could be used as a starting point for this memorization process, with ICM, OCO and MICA positions becoming progressively more attached to specific move sequences in "muscle memory".

Finally, some arbitrariness is still present in the described steps to the 2mo-4mo transition. Namely, there are often multiple good cycles available to be formed, and the last move of the cycle rotation (M or M') can be arbitrarily chosen in the cases where ICM and OCO are the diagonally opposite edges of step 2.

Future proposals could further restrict this freedom in order to go beyond 4b, worrying about "good 4c" cases even before 4a. Therefore, I hope the ideas on the present work can contribute with the understanding of the last six edges behaviour, and perhaps sparkle further contributions that could open the doors to one-look LSE.

TL;DR - (EOLR) How to intuitively get a good arrow from 2 misoriented edges

Definitions



Procedure


  1. Setting up: keeping white or yellow on top, adjust the U face setting up an F- or B-cycle such that:
    1. The cycle is composed of three oriented edges and a single misoriented one.
    2. Adjacent to the in-cycle misoriented (ICM) edge, there must be an oriented edge that matches either the ICM or the out-cycle oriented edge (OCO). Call this edge MICA (matching in-cycle adjacent).
  2. M or M’: raise the cycle you set up, i.e. do an M’ move if it was the F-cycle or an M move if it was the B-cycle.
  3. U or U’: place either the MICA or the ICM (whichever was outside the M slice) diagonally opposite to the OCO.
  4. M or M’: reposition the M slice to keep the ICM and OCO in different layers, one in U, the other in D.

Multiple valid cycles can sometimes be found in step 1, but also none, in some rare and specific situations, even if EOLR can still be solved. Thus, further ideas and improvements are appreciated.

A. Example solves

Case 1 (3D view)

Good B-cycle selected: ICM=UL MICA=UB, OCO=UF; Matching pair is MICA-OCO

Case 2 (3D view)

Good B-cycle selected: ICM=DB, MICA=UR, OCO=UF; Matching pair is MICA-OCO

Case 3 (3D view)

Good F-cycle selected: ICM=UF, MICA=UR, OCO=DB; Matching pair is MICA-ICM

Case 4 (3D view)

Good F-cycle selected: ICM=UF, MICA=UL, OCO=UB; Matching pair is MICA-OCO

Case 5 (3D view)

Good F-cycle selected: ICM=UL, MICA=UF, OCO=DB; Matching pair is MICA-ICM

Case 6 (3D view)

Good B-cycle selected: ICM=UB, MICA=UR, OCO=DF; Matching pair is MICA-OCO