Pairing up the Edges (Dedges)

There are numerous ways in one could pair up the edges to form double-edges or "dedges". However, in speedcubing, two methods predominate, "2-pair", and "6-pair". In the 2-pair method, 1 pair is made by slicing through the centres, and then a 2nd pair is formed when the slice is undone. In the 6-pair method, 3 pairs are made by slicing through the centres, and 3 more are made when the slice is undone, a total of 6 pairs. It will therefore be necessary to perform more slices using the 2-pair method, but is much less likely to present you with awkward cases as the 6-pair method is prone to do. However, I believe that if you have an efficient way of dealing with the awkward cases, the 6-pair method is more likely to lead to very fast times; it is the method I use and the method I present to you in this section.

The First Iteration

When pairing up the edges, divide the cube up into 3 sections, a working area and two storage areas. As the names suggest, you will form the pairs in the working area, and then store them in the storage areas. Opinion is divided on how best to divide up the cube, some cubers prefer to hold the cube so their storage areas are the R and L layers of the cube, others, like myself, prefer to form the dedges in the u and d slices and store them in the U and D layers. I will show you how the method works when the storage layers are in U and D, and once you get the hang of the general idea you can experiment with holding the cube in both orientations.

 

Edges will be formed in the Working area (the white sections of the cube), and stored in the Storage areas (the grey sections). Of course, the orientation of the cube doesn't matter initially, so the Centres could be any colour, it's not necessary to always to hold a certain colour on top.

To help make the explanation a bit clearer, I shall label the edge pieces. The following diagram shows how the pieces are labelled (not a standard labelling, it has just been devised for this webpage), and these labels are referred to throughout the following text.

Our starting point is at edge 1. We must then search the cube for the other edge piece which has the same colours as edge 1, and we need to replace the edge at position A with this other edge piece. Hopefully it will be somewhere in the U or D layers, which makes life simple for us. If it is, there are 4 possible scenarios which can occur (after positioning the U or D layer correctly). You will have to get used to each case, and be able to recognise it fast and execute the moves without hesitation.

R U' R'	U' F' U F	D F D' F'	R' D R

Once you have inserted into position A the edge that corresponds to edge 1, you will have a new edge in position 2. You now must repeat the process, this time finding the edge which corresponds to edge 2, and insert it into position B. And once you have completed that, find the edge that corresponds to the now new edge 3, and insert that into position C. At this point, you will have performed 3 insertions (unless you already encountered some awkward cases, more about that later), and you will have formed 3 staggered edge pairs. A slice turn will now make 3 dedges, 1A, 2B, and 3C.

Now that you have formed 3 dedges, you can use the same technique to form another 3 on the "way back". In the working area, there is now only 1 dedge that isn't formed (4X), and you should be concentrating on the edge at position 4. The idea is to now to store the formed dedges in the U and D layers, whilst forming the staggered dedges as we did before. So we find the edge that corresponds to the edge at position 4, and we insert it into position C of the dedge 3C. We then look at the new edge in position 3 and insert the other corresponding edge into position B, at the same time storing the dedge 2B. And finally we look at the new edge in position 2, and insert the corresponding edge into position A. Finally, we slice back to restore the centres, and form another 3 dedges. In total, we have completed 6 dedges, and finished the First Iteration.

The Second Iteration

After completing 6 dedges, the centres are restored and for all intents and purposes you are back where you started, albeit 6 dedges better off. You can now complete the next 4 dedges in the same way as the first 6. The main difference here is that you only need to form 2 staggered dedges before you slice through the centres, thus on the way back you only form another 2 staggered dedges, which is 4 in total.

Referring back to the initial labels, you would therefore want to look at the edge in position 1, and insert the corresponding edge into position A. Then look at 2 and insert into B. Perform a slice turn to form 1A and 2B. Now look at edge 3, and insert the corresponding edge into position B of dedge 2B. This stores dedge 2B safely at the same time. Finally, look at the new edge 2 and insert the corresponding edge into position A of the dedge 1A. Slice back to restore the centres. 4 more dedges completed, 10 in total!

The Final Two Dedges

It may seem trivial to say, but after 10 dedges have been completed, there are only 2 possible scenarios for the remaining 2 dedges. They may be correctly formed, in which case you have completed all the dedges. Or they may be incorrect, in which case you will have to fix them with a small trick.

solved	d (R U R' F R' F' R) d'	L2 d2 (R U R' F R' F' R) d2 L2

The basic idea is to get a situation where the X's are in one layer, and the Y's are in the other. You then slice through the centres to replace one of the Y's with the other. Next, you apply a move to flip over the dedge (the move in brackets), which exchanges the X for a Y. Then, when you slice back, the Y will line up with the other Y, and the X will line up with the other X, and the dedges will be solved.

This is the general 6-pair method and how it works. It's likely however that you will encounter some awkward cases which prevent you from being able to solve the dedges in this standard manner every time, but these are easily resolved if you know how. Some special tricks for dealing with these cases are shown on the next page

Solving the Final Two Centres + Special Tricks

Case 0	Case 1	Case 3	Case 4
solved	r U r'	r U' r2' F2 r	r U2' r2' F2 r
Case 2a	Case 2b	Case 2c	Case 2d
r U2' r'	r U' r' U r U2' r'	r U' r' F r U2' r'	r U' r' F U' r U2 r'

Case 0
This is what you are aiming for. In the diagrams, Yellow is on R, Red is on U, and Green on F.

Cases 1 and 3
When you see blocks of 3, align the U and F faces so that lone sticker is in a position which corresponds to the "middle block of the L" on the other face. For example, in Case 1 the middle block of the Red L is at the Ufr, the "South-East" corner of the U-Centre Block, so we position the lone red Centre sticker at Fdr, the "South-East" corner of the F-Centre Block. In Case 1, we can do a simple insertion, but in Case 3, where we have to correct 3 pieces, we first do a trick to get to Case 2a, and then we solve it in the same manner.

Case 4
It is pretty rare to have to completely swap over the 2 Centre Blocks, especially if you are careful to avoid it while solving the 4th Centre. However, if it does come up, it is straight forward to fix. First, transform it into Case 2a, and then solve in the same manner.

Case 2a
A simple overlap insert. Just make sure you overlap rows of the same colour! (see diagram)

Cases 2b and 2c
The trick with these cases is to get the rows of 2 "pointing" at opposite colours. For example, in Case 2b the row of 2 Reds is "pointing" at a lone Green on the U face. You can then always use a move sequence of the form a b' a', to transform it into Case 2a, and solve in the same way.

Case 2d
The trick here is to make both faces look the same. In the diagram the Red stickers on both faces point North-West to South-East and the Green from North-East to South-West. As long both faces look the same you can use a move sequence of the form a b' a' (see Cases 2b and 2c) to transform into Case 2a.

Swapping Two Opposite Centres

At some point in your 4x4x4 solving you will inevitably make a mistake with your Centres, such as transposing two Opposite Centres. This can happen if you are used to placing your last 4 Centres in a certain order, but accidentally hold the wrong coloured Centre on the R face. It is possible that you won't notice this until you are well into solving the reduced 3x3x3 stage, because you will find you cannot solve it! There is an easy fix however, so don't panic if it happens to you!

u2 R2 L2 u2 (swapping L and R Centres)

Solving the Reduced 3x3x3 Cube

So, you might wonder, what exactly have we achieved by this point? We've made the Centre Blocks, and paired up the dedges, but so what? The cube is still a mess! But at this stage the cube is an ordered mess. In creating the Centre Blocks, we have made the equivalent of a Centre piece on the 3x3x3, and by pairing up the dedges we have created the equivalent of 12 single edges, as on a 3x3x3. Add in to the mix the 8 corners, and we have the equivalent of a 3x3x3 cube on our 4x4x4 cube! It can be solved in exactly the same way, and by making sure to only use outer layer turns you can't possible destroy the Centres or Dedges. I assume you can already solve the 3x3x3 if you are learning this 4x4x4 method, but if you need some help with that then please

However, this is called Rubik's Revenge, and Rubik still has some tricks up his sleeve even at this point. Because it is only a simulation of a 3x3x3, there are two problems which you can run into, which would be impossible to solve on a standard 3x3x3 Rubik's Cube. These are called Parities.

Orientation Parity

After solving the First 3 Layers of your 4x4x4 (the F2L on the simulated 3x3x3 cube), this would be the first problem you would notice, if you encounter it, and it would be the first thing you would solve. If you have parity at this stage you will notice that an odd number of LL edges are incorrectly oriented. To fix parity, you have to flip over a single dedge. With the 3x3x3 reduction method of solving the 4x4x4 you have a 50% chance of encountering the OLL parity problem.

l2 B2 l U2 l U2 (x') U2 l U2 l' U2 l U2 l2 U2 (flips UF dedge) r2 B2 r' U2 r' U2 (x') U2 r' U2 r U2 r' U2 r2 U2 (flips UF dedge, mirror of first alg) r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 (flips UF dedge)

This OLL Parity fix also affects other pieces in the LL, but since it is the first step in the solution to the LL, it doesn't matter. Once you have corrected the OLL parity, you would then solve the OLL as normal.

Permutation Parity

The Permutation Parity Problem is much harder to recognise than the Orientation Parity Problem, and is only really apparent after you have completed the OLL. When the parity of the permutation is odd, it is impossible to solve using standard 3x3x3 PLL algorithms. Hence the way to recognise it is if you "don't" recognise it! To fix parity, it is easiest to swap a pair of dedges. With the 3x3x3 reduction method of solving the 4x4x4 you have a 50% chance of encountering the PLL parity problem.

d2 r2 U2 r2' R2 U2 r2 d2 (swaps UF with UB) r2 R2' U2 r2 R2' u2 r2 R2' u2 (swaps UF with UB)

Once you have corrected the PLL parity, you would then solve the PLL and thus the whole cube as normal.

Final Thoughts

The parity problems can be big time-killers in a speedsolve. In 1/4 solves you won't encounter a single parity problem, but in another 1/4 solves you should expect to have to fix both problems. And in 1/2 of your solves you will have to fix 1 problem only. The most efficient way to deal with these is still up for debate, but in the

However, if you have got this far, and have been successful, congratulations on solving your Rubik's Revenge!