Edges - Algorithms & Concepts
Ok so now you have an idea of how to come up with an order to memorize and solve in. I gave arguments for my way, but of course my way is not the only way to do this.
When solving the corners, be very careful to only use supercube safe algs (i.e. don’t twist a center group!! If you do your whole solve will be a DNF!).
Ok so now onto solving the edges. I’ll break this up into an overview of commutators, then list a beginner, intermediate, and advanced method for using them to solve the cube blindfolded.
EDGES OVERVIEW:
What we are going to do is to use nothing but intuitive commutators for this step. I don’t use a single "alg" for this step at all. The purpose is to understand exactly and 100% completely how you are moving only 3 edges at a time.
As you may have guessed I do this step using three cycles, and by doing this you will solve 2 pieces at each step.
The idea of an intuitive commutator works like this. There are certain requirements to start, and the process is always the same once you have met those requirements. So let’s start with what has to be true in order for a commutator to exist to solve a situation.
1) Two of the three pieces to be cycled must be on the same slice.
There are 12 slices on a 4x4x4 cube, and two pieces must be on the same slice in order for you to even consider thinking about a commutator to use.
2) One of the three pieces to be cycled must NOT be on the slice as the other two.
This is very important. So much so that this edge is given a name, the "lone edge".
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Ok so now we have the requirements, let’s talk about the process of coming up with a commutator.
Say we have the cycle (bUL->dFL->fUR). Now think about how they look on your cube. You will do a lot of mental picturing of where these pieces are in a blindfolded solve, so preferrably do this without a 4x4 cube handy.
Ok some questions.
1) Are there two pieces in the same slice?
Yes there are, in two different ways. bUL and dFL are both on the L face. Also bUL and fUR are both on the U face.
2) For each of the ways to have two pieces on one slice, does there exist a "lone edge"?
Yes. If bUL and dFL are on the same L slice then fUR is the lone edge not touching L. If bUL and fUR are the two pieces in the U slice then dFL is the lone edge not touching U.
Ok now I have to introduce a concept called "interchangeability". I used to use a weird way to describe this concept, but after talking with Doug Li he hit on this phrase and now it is the one I use too.
Not only do you need two edges in the same slice, but they must be interchangeable. This means that by turning that slice, you can make one of the edges occupy the spot the other originally occupied.
As an example, for the cycle above we have two ways to put two edges on the same slice and choose a lone edge. But which way do we choose during a blindfolded solve and why?
I choose to have the two edges in the U layer rather than on the L layer. This is because the two edges in the U layer are interchangeable and ones in the L layer are not.
Let’s look at the two possibilities:
L layer edges: bUL and dFL. Can you do an L move and put bUL into dFL’s place? No. Can you do an L move and put dFL into bUL’s place? No. So these edges are not interchangeable.
U layer edges: bUL and fUR. Can you do a U move to put bUL into fUR’s place? Yes, U2. Can you do a U move to put fUR into bUL’s place? Yes, U2. So these two edges ARE interchangeable.
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Now on to the cycling. Remember our cycle is bUL->dFL->fUR and we have chosen our two edges to the ones in the U layer because they are interchangeable. This leaves dFL to be our lone edge.
Ok now here is the process.
Step 1) Do some move that places the lone edge into the spot of the edge it cycles to WITHOUT messing up anything else on the main slice.
In terms of our example look at it like this. The U layer is our "main slice" because it is the one that the two interchangeable edges are on. The lone edge is dFL. Now look back at our cycle, where does dFL move to? If the cycle again is bUL->dFL->fUR then the dFL edge goes to fUR.
Here’s the key though, come up with some move that moves dFL to fUR and does not touch ANY OTHER PIECE IN THE U LAYER. This is they key that makes this work.
So come up with a move, then check with mine.
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My move to place dFL into fUR without disturing the U layer is R’ d R. If your move is different, then as long as it did not disturb any other piece in the U layer other than fUR then you are ok.
Now we are onto step 2
Step 2) Interchange the unused edge on the main slice to the "action spot".
Definition of terms based on our example. The "action spot" is the part of the U layer you just moved the lone edge to. For our example the "action spot" is fUR. This is literally where all the cycling and all the "action" is taking place.
So of our two U layer edges we dealt with fUR but not with bUL yet. So what was the move to interchange bUL and fUR? Right, U2.
So do U2 now.
Step 3) Do the inverse of the move you did to place the lone edge into the action spot.
Basically take the action spot edge out and put it back in the lone edge’s original spot. For the purpose of our example take the edge currently in fUR and place it in dFL in the EXACT REVERSE way you placed dFL into fUR. If you used my move to place dFL into fUR as R’ d R then now do R’ d’ R to place the action spot edge (fUR) into the lone edge’s original spot (dFL).
Theory: What you are now doing is the A’ part of the basic commutator form A B A’ B’. Part A so far has been our move to place the lone edge into the action spot. B was our move to interchange the unused main slice edge to place it into the action spot. Now we are to A’ in the commutator move which is to undo the move we used for the A part of the commutator. Understanding the theory of what you are doing is very important, but here’s a catch. During an actual speed bindfolded solve, shut your brain off at this point and just mindlessly undo A. You don’t care what pieces are moving where and why, you will know already that just moving the action spot to the lone edge spot will be part of creating a complete cycle.
Step 4) Undo the move you did to interchange the unused main slice edge into the action spot.
For our example, we used U2 here, so now undo with U2 again.
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You have now done the commutator R’ d R U2 R’ d’ R U2 which cycles the edges as bUL->dFL->fUR. Now if I had just given you the alg R’ d R U2 R’ d’ R U2 and said learn this, you would have thought that blindfolded cubing is all about learning and memorizing algorithms. But it’s not! We never once thought "Ok I need to use the R’ d R U2 R’ d’ R U2 algorithm since this is the bUL->dFL->fUR cycle" all we did was to intuit how to move those edges.
So here is the process:
1) Find two edges in the same layer.
1a) check to see if these edges are interchangeable
2) Find a lone edge that is not on the same slice as the other two *interchangeable* edges
3) Call the slice with the two pieces your "main slice" and the other edge the "lone edge".
4) Now move the lone edge to the location it goes to based on the cycle. This location will be one of the main slice edges. This location becomes known as the "action spot"
5) Do the interchanging move that moves the unused main slice edge into the action spot.
6) Undo the move you used to place the lone edge in the action spot. You can think of this also as moving the action spot edge to the lone edge spot.
7) Undo the interchanging move you did to place the unused main slice edge into the action spot.
Now that you know how to do it, let’s get onto the only sort of quasi memorizing that you will have to do to get faster at blindfolded solving a larger cube.
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Beginner method to blindfolded solve a larger cube using freestyle cycling
Ok so now we define the equivalent of algorithms to blindfolded solving. You don’t use algorithms here you use "commutator types". This is basically an algorithm, but remember you do these commutators on any slices, from any angle. This is because you aren’t rotating or reflecting an alg. You’re just using the same process as defined above, but the moves will be fairly similar.
There are two commutator types you MUST use for the beginner level method of free style cycling. Any less and I consider you to be using a method that is unnecessarily hard on yourself and will require a lot of setup turns, which are hard to handle during a solve if you must do a lot of them in a way that is unstructured. Stefan’s method is an exception, since the setup turns are very structured and done with a system behind them. Setup turns in freestyle cycling are as much fun as getting the flu, so avoid them at absolutely all costs.
You must learn these two commutator types if you want to be a good beginner solver using freestyle commutators:
1) The Direct Insert
2) The Drop and Catch
What I do for these is to give an example that is on the U layer to represent each commutator type, but remember there are 24 ways to rotate your cube from each of these examples, and you can also reflect the representative case and rotate it 24 ways. So by learning one commutator type you will know at a MINIMUM 48 algorithms to use during blindfolded solving! Yes that’s not an exaggeration, you will be able to actively use 48 algorithms simply by knowing the process above and the name Direct Insert for example. Also you can move the interchangeable edges around on the main slice and create even more algorithms! So learning these two types teaches you a huge number of algorithms!
The commutator types:
Direct Insert
The representative cycle is rDF->rUF->fUL
The parts
A = F d F’
B = U’
This is my way or writing a commutator type. Here is how to read it. Imagine the cycle rDF->rUF->fUL and see why I call it a direct insert? Let’s find the required parts of what we need to use our process.
Where is the slice with two interchangeable edges? We can choose the inner r slice which has the interchangeable edges rDF and rUF (interchange them by doing eithe r or r’) or we can choose the U slice with two interchangeable edges rUF and fUL. This commutator type chooses to have the U layer edges as our interchangeable edges and the rDF edge as the lone edge.
Ok now that you know the setup look at the A part of the ABA’B’ required form the commutator must take. See how it directly inserts the rDF edge into the spot right above it (the rUF spot)? This shoots the edge directly above it into the U layer. So it is a direct way to insert the edge into the U layer. Think of those pieces as in a line, and you ride the elevator all the way to the top floor.
Now if you do the move F d F’ to place the rDF edge into the newly defined action spot of rUF now what? You have to move the unused edge in the main slice to the action spot. This you do with U’ and that is the B part of the commutator type written above. Ok now you know the theory already, we will do the inverse of the A move next to move the action spot to the lone edge spot (DON’T JUST DO r’!!!! This affects the U layer! Remember you must do each of these steps so as not to affect the main slice in any way other than at the action spot!). Once you have moved the action spot back to the lone edge spot, undo the interchanging move by moving the action spot back over to the original unused spot. So do U here.
So though I only list the A and B parts of an algorithm you know to put them together as ABA’B’.
Ok now here is the catch. Is the cycle dFR->dBR->rUB a direct insert? It’s certainly not going to use the U layer much if at all.
The answer is yes it is. The main slice will be the B slice, with interchangeable edges dBR and rUB. And the dFR edge is in a DIRECT LINE with the dBR edge and ready to insert onto the B slice.
So this situation you would use R f R’ as your insert move, then B’ as your interchange move, then undo the insert move with R f’ R’ then undo the interchange move with B and you have done a direct insert on the B slice.
I’ll leave you to figure out how the following are direct insert cases:
1) cycle fUR->fDR->lDF
2) cycle uFR->uFL->fDL
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Ok now onto our second commutator type for a beginner’s approach to blindfolded solving using free style commutators.
The second type is called a "Drop and Catch".
Drop and Catch
cycle: dFL->fUR->bUL
A = R’ d R
B = U2
And this should look familar, it’s the first example we did to describe the commutator mental process. Now why is it called a drop and catch? Remember our main slice is the U slice and the lone edge is the dFL spot. See how we insert the lone edge into the U layer though? We drop it down with R’, then we slide the dFL edge over with the d turn. Consider the lone edge as a fast ball that was moving very quickly, and we dropped the R slice down to catch it before it went by. Then bring the edge back up to the U layer with R.
Now I leave it to you to see how these situations are also a drop and catch:
1) cycle dFR->bDR->rUB (think about the fast ball and trying to catch it. I’ve moved the edges around a bit so it’s not exactly like the first example, but in spirit you still drop a face, then catch the lone edge)
2) cycle dBL->rDF->bDR
3) cycle uFL->bUL->rUB (there are two ways to drop and catch here, find both. One of them will involved doing the drop and catch as BAB’A’ rather than ABA’B’)
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Inverting a freestyle commutator
Here’s is how to invert a freestyle commutator. I try to always insert the lone edge into the main slice first, so I am always doing my A move first. However, sometimes this isn’t possible and you need to invert the commutator.
So think about ABA’B’ and what is the inverse? You get BAB’A’ which is the commutator done with the B first! In terms of your thought process all you do is to do the interchange move first, then follow through with the rest.
Say you have the cycle dFL->bUL->fUR
This looks familar right? But now the lone edge goes to the OTHER spot in the main slice, not fUR. Instead of treating the two pieces on the L slice as your main 2 edges and doing a setup move to make them interchangeable, just do the inverse of the drop and catch.
So first I would do U2 which moves the bUL spot to the fUR spot (which will be our action spot). Now I insert the lone edge into the action spot. Then I undo the move I did to place bUL into the action spot. Then I undo the move I did to insert the edge and I finish this way.
***USEFUL TIP*** when doing a commutator when you are actually blindfolded remember this tip. If you start with a main slice move, you end with an insert move. If you start with an insert move, you end with a main slice move.
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I’ll stop here, since I think most of the people reading this tutorial are just interested in a way to get started and not about the more advanced stuff. I’ll post another post later that has all the commutator types I consider essential for an intermediate approach, and also every single commutator type I use for the more advanced section. If you count in a very theoretical way I think I use around 20 types. If you just count by names and the spirit of what a situation looks like then I think I use around 10 types. That’s just a guess, I’ll come up with more concrete numbers when I actually do that post.
Hope this helps. Please post any comments!
Chris
Need I say more??? I've wanted this FOREVER!!! hehe...
