Speedsolving Posts: A Huge Cube Method

I really put a lot of thought into making a method for every part that would scale well - remain easy even with extremely large cubes. And when I say extremely large, this is the method I solved the 111x111x111 with. It's probably still useful on smaller huge cubes (like 20x20x20) but when you can see a whole face at a time, and count layers, some parts can be done quite a bit faster.

From http://www.speedsolving.com/forum/showthread.php?42084:

First, some basics. I solved most faces in two steps: "cleaning" and finishing up. The idea of cleaning a face is to solve as many pieces as possible as quickly as possible. After that step, which typically solves 50% or more of the face, I can go through the longer step of putting each of the remaining pieces of that color into the center, one at a time or in blocks. As for notation, I'll be using something like SiGN (xR means turn the xth layer on the R face, or the first layer if x is absent), except that I'll also be using lowercase p and q as variables to better describe classes of commutators.

1) Solve the edges. I do edges first for large computer cubes because reduction (centers first) scales badly, and because having solved edges actually lets me use them to keep track what part of the puzzle I'm working on. A lot of the time I do a pR' move or something to start working on a row of one center, and the incorrect edge reminds me which row it is. With so many layers it's impossible to do it by counting or visual inspection alone. Anyway, for the first 9 edges I did basically the same thing as with other cubes. For the last three edges I used moves like pR U2 pR' (for the 10th edge only), pR2 U2 pR U2 pR2, and pR' F U' R F' U pR. Instead of solving one orbit at a time like on my keyboard solves, I went for an edge at a time and solved pieces in the same position in groups. For instance on the last 3 edges there are only 6 possible positions for each piece, so with 109 pieces in each edge there are a lot of pieces in the same position which can be solved in the same way, all at once, by replacing a pR move with a whole bunch of slice moves along the same layer. This took about an hour in total.

2) Clean the first center. I used white. Hold the white face on U, and then for each row (perpendicular to the F face), we do something like this: move white centers out of the way on the corresponding row on F using pU moves, then qR', then solve as many white centers into that layer as possible using only pU-axis moves, then qR. By "solve as many white centers ... as possible" I mean I basically look at that row on R, B, and L, and for every white center I see, move that layer the right amount. Note that these rows are vertical, which is something I did because of the way Iso is set up (when zoomed in, it's easy to scroll through a vertical row on F or R using just the up/down arrow keys). You could do any of this from a different angle if you wanted. A typical layer might be something like (2U 4U 7U ...) 3R' y (3U 6U 8U ...) y (U2 4U2 8U2 ...) y (2U' 3U' 4U' ...) y 3R. Each center on white corresponds to 5 stickers, and a white center on any of those 5 stickers means the center gets solved. I think roughly 60% of the center ends up solved.

3) Finish the first center. Now I can hold the center on F and look at R. If I see a white center anywhere, I can solve it using pF qU' pF' qU. Four-move commutators are fast! However, I can't just do that randomly, but instead, I have to do it so that the center moves into a place on the white center that didn't have a solved center already. To check if there is room I do qU' qU, and if there isn't, I rotate the white center and repeat as necessary. You'll see this type of checking often as I continue. Note that this requires centers to be adjacent, and since we're solving from all five, we have to swap two adjacent centers. On the 5x5x5 the algorithm for this is 2R U2 2R' 2L' U2 2L U 3R U' 3R', which you then have to properly undo later.

4) Clean the second center. I used yellow. This is the same idea as the first center - hold yellow on U (so, white on D) and proceed row by row. However, instead of using pR' moves to move a row onto the F face, I used pR U2 pR' moves. This does solve the layers in a different order, but that's okay. For the middle layer I used U pR U' pR'. Each center on yellow again corresponds to 5 stickers (any yellow piece in one of those 5 places means that piece gets solved), and I think roughly 67% of the center ends up solved after this.

5) Finish the second center. Again I swap yellow with another center, so I can hold yellow on F, white on U, and the solved center on R. The commutator this time is pF2 qU' pF2 qU (or pF2 qU pF2 qU' for the mirror). Since I can't just do x rotations and keep going, I have to do some extra center swaps to bring each of the four non-white/yellow centers into the right place. There was also plenty of the qU' qU moves to check if there's room to put a piece.

6) Clean the third center. I used orange (for visibility). Place orange on U and white/yellow on L/R. Now we can do pU2 moves but not pU moves. So I do something like the white center cleaning, where I do pU2 moves to move orange centers out of the way, then a qR' move to bring the row down, then do more pU2 moves to solve as many orange centers as possible, then a qR move. A typical row clean might look something like (2U2 4U2 7U2 ...) 3R' y2 (2U2 3U2 4U2 ...) y2 3R. Each center on orange corresponds to only 3 stickers, solving about 57% of the center.

7) Solve the third center. This is just like what I do on yellow, with the pF2 qU' pF2 qU moves, since they keep U and D solved, and again I use plenty of qU' qU moves. In fact, there are fewer center swaps than last time, since the D center can stay where it is.

8) Clean the fourth center. As a general rule you want this to be adjacent to the third center, so I chose green (since it was brighter). Unfortunately the cleaning setup I use only really lets us clean a center opposite of the third one, so we place orange on D and green on F, then swap F and U, and begin. The cleaning itself is just like the third center except, again, we use pR U2 pR' rather than pR'. Each center on green corresponds to 3 stickers, solving about 70% of the center.

9) Solve the fourth center. Things got pretty messy here and it took a lot longer than I expected; most of that was probably because I can't use 4-move commutators any more and thus had to switch to 8-move ones. I held green on F and orange on D, and then did commutators like [pU2, qR U qR']. In fact I did all of the pU2 moves ahead of time, so that for each row I could immediately do a qR move for each piece I wanted to solve (using pU2 pu2 to check), and then solve many piece in that row at once. The problem, apart from the large number of qR moves, is that I couldn't stray too far from U, because I had to do U turns. So I ended up having to stay on the top half of F, and doing a lot of F and B moves to be able to solve things. Oh yeah, and once I got rid of all the green pieces on B, I had to swap U and B and do it again.

10) Solve the last two centers. No cleaning here - I couldn't see any way to quickly solve most of the center. I held the two centers on F and R, and solved one vertical row of R at a time. The commutator looks something like [pU', R' qU' R]. Basically, I would do a pU' for every center in that row that I could solve, then do the R' qU' R, undo the pU moves, and finish the commutator. Then rotate F and go again - in at most 4 separate block commutators I would solve all pieces in that row. (With one caveat - if the x-center closer to D is unsolved, this commutator won't work on it. When that happened I had to do it individually.) After each finished row I did a qF2 move to get it out of the way and make the next row easy to see; since I had blue on R, I ended up with a whole bunch of green rows on that face. Since I had to be in view of the F-R edge to do R tuns, I only did the first half of the blue face before undoing all those green rows, turning that center by 180 degrees, and doing it again.

And that's it! It might sound kind of complicated, but each step is pretty straightforward once you know what's going on, and I spent the vast majority of my time doing the same kind of thing over and over. The low move/piece count was mostly due to the cleaning stages, where I could solve a ton of pieces in one or two moves each. Those stages would have an absurdly low movecount in axial turn metric / snyder metric, with something like 400 moves solving 5000+ pieces. And in fact, you can do a cleaning stage on the same center more than once - just turn the center to a different orientation and continue. With four of these cleaning stages you should be able to solve every piece on the affected centers. I didn't do this because the returns do diminish, and so I don't think it's worth it to clean more than once for any individual center. Maybe for the fourth center it's worth it though! 

Speedsolving Posts: Parity Errors

I'm going to be posting some of my longer posts on speedsolving.com here. I think I've written some very useful things down there, but it's a kinda big forum (and I've made a lot of posts!) so it's easy for stuff to get lost. The plan is to copy some of them here, with a link to the original post (for those who want to view the thread). Here goes...

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From http://www.speedsolving.com/forum/showthread.php?p=852756:

In speedcubing there are two things we think of as parity: "reduction parity" and mathematical parity. Some things we call parity fall into both categories, but others only fall into one of them, and you will often see people disagree on whether certain things count as parity, because they disagree that both of these two definitions are valid.

Reduction parity occurs when you try to reduce the puzzle so it can be solved by a constrained set of moves, putting it into some subset of the positions. However, you can often reach a position which seems like it is in your subset, but which is actually not, and to solve the puzzle you have to briefly go outside your constrained set of moves to bring the puzzle back into the subset you want. Typically the number of positions you can encounter is some small multiple of the number of positions you expect. The obvious example is PLL parity in 4x4x4: all the centers and edges are properly paired, so you expect to be able to finish the puzzle with only outer layer turns, but this isn't quite possible. OLL parity falls under this definition too (so the reduced 4x4x4 has four times as many positions as you would expect). Square-1 parity also falls under this definition - your constrained set of moves are any moves that keep the puzzle in cubeshape. BLD parity is not of this type (the solver has not reduced the puzzle).

Mathematical parity is based on the idea of the mathematical definition of an odd permutation. Basically, at least one orbit of pieces has an odd permutation, and thus cannot be solved with just 3-cycles. You see this type of parity crop up in blindfolded solves, because blindfolded solvers attempt to solve most or all of the puzzle with 3-cycles and thus an odd permutation is very noticeable. OLL parity on 4x4x4 and Square-1 parity can also be thought of as this way, as they originate from some kind of 2-cycle. PLL parity on 4x4x4 is not of this type (it can be solved with 3-cycles).

Pixeling, Or, Putting Hearts on Your 7x7x7

Pixeling is a technique that I discovered to make certain types of pretty patterns on NxNxN cubes. It affects the centers, only, in a large series of 3-cycles. Specifically pixeling is used to generate patterns where a certain shape of centers is moved in two 3-cycles, from (say) the U->R->F faces and D->L->B faces, in a very efficient way. An example is the hearts picture at left, which can be created with just 11 block turns (i.e. turns of a group of adjacent slices all in the same direction)! If you ask me that's ridiculously efficient.

Here's the motivation (with pictures below). You'll need to try these patterns out to understand where pixeling comes from. First we start with the basic center pattern on the 3x3:
M E' M' E.
This just moves the centers around. We can also do this on the 5x5 (or any NxNxN cube, really):
2-4r' 2-4u 2-4r 2-4u'.
But on the 5x5 there are related patterns. The following show that you can move blocks of centers around (or a single center), and that the blocks don't even need to be contiguous:
2-3r' 2-3u 2-3r 2-3u'
(2R' 2L) (2U 2D') (2R 2L') (2U' 2D).



You can basically now make any pattern we want using a series of these block maneuvers, and the efficiency on those patterns is not bad at all. But we can do better. Try this maneuver on a 5x5, which creates the staircase pattern to the left:
2R' 4U 3R' 3U 4R' 2U 2-4r 2-4u'
This is the most introductory example of pixeling. Do you see what is happening there? It's a bit tricky, but try to keep track. Every time we do a U turn, we are basically storing a line of three centers on the L-face. Together, those lines make the entire center (and the shape that we want to pixel in, which is inside it), which is done when we make the 2U move. Then we simply return the vertical slices to their original position and move all the horizontal slices back. This can also be done using the vertical R slices to store the center lines in the D face, of course, and some patterns turn out to be more efficient one way than the other, so it's important to try both ways.


We're not limited to just R' turns, though - sometimes we want to use R slices to remove a center from our shape. To the left is an arrow shape for the 7x7 that can be created like this:
2-6r' 4U 2R 2L' 3U 3R 3L' 2U 2-3d' 4R 2-6u'
Notice that before every time we do a turn of a horizontal slice in the U direction, we set up the R slices so that only the ones which contain the centers we want in our shape are misaligned. Also, this time we didn't do the U slices in order, but this is also fine; the pixeling maneuver still works.

The end result is that there are many, many ways to pixel any given shape. It will take a bit of trial and error to do a shape efficiently, but the efficiency can be really amazing. If you're making a shape yourself, remember to try both directions (vertical and horizontal slices to set up the shape) and a few different orders of slices. If one of those cycles in the wrong order that you want, you can just invert the pattern. Finally, let's return to that heart pattern. Here's the best we can do for the heart pattern with blocks and with pixeling, respectively:
Blocks: (2R' 2L 3-4u 2R 2L' 3-4u') (3R' 3L 2-5u 3R 3L' 2-5u') (4R' 3-6u 4R 3-6u') (16 moves)
Pixeling: (3-4u 2R' 2L 2-3d' 4R' 2D 2U 3R' 3L 2-5u' 2-6r)' (11 moves)

Happy patterning! Thanks go to Werner Randelshofer for the cube animation applets and to Joël van Noort for his NxN ImageCube (which I used to generate the images you see here).

How to Solve 6x6+ (on hi-games)

Let's start by saying that I don't really like the "grip shift" controls. Thus, the way I solve big cubes is designed to be somewhat fast (although I'm sure there are faster ways out there) while changing the grip as few times as possible. As it turns out, reduction is the best method for this.

Centers

Obviously, centers require lots of face turns, but also lots of slice turns. So how can we solve them with a small number of grip shifts? As it turns out, we can do a pretty decent job using U, F, D, B, and any r and l slices. Only build and insert rows on the x axis (faces U, F, D, and B), and although this uses a lot of cube rotations it isn't too restrictive. Besides, you can only see U and F anyway. So then the question is what slices we need. Here's how I do it:
6x6 and 7x7: shift both right and left in once. Now you can do all of the usual moves (3r, 3R, 2r, 3l, 3L, 2l) with each hand.
8x8 and 9x9: shift right in twice and left in once. The left hand can do 2l, 3l, 3L and the right hand can do 4r, 3r, 4R.
10x10: shift right in three times and left in once. The left hand can do 2l, 3l, 3L and the right hand can do 5r, 4r, 5R. Note that 4R is actually not possible (except by doing 4r y2 3l'), so you should try to work around this when building centers.

Note that in the cases where the right and left hands are asymmetrical, I really suggest pairing from the inside out, since slice moves are not always possible or convenient. The goal is to avoid cube rotations so you can spend all your time actually doing moves. As with other large cubes, by the way, I really suggest pairing up 1x(n-1) rows of centers, and putting them together as soon as they are done.

Edges

My method for edges is worse, but it still works pretty acceptably. The way I do this is to use the D slice to do slice moves and pair up edges on the E slice, since I tend to use the other slices to move edges around. Currently I use WO for moving the D slice in (up) and SL for moving it out (down), but you can use anything you want. Then I pair up edges multislice-style: I start by pairing up the innermost pieces, and then move out. Note that it's also possible to just use the L slice (and pair up edges on M), and use only RUDFB moves to insert edges, but personally this doesn't feel natural to me so I don't use it.

My method of pairing up 5x5-style edges is freeslice (the bigcubes.com) style, which means I pair up one edge group (of 3 pieces) at a time and don't fix centers until I have eight edges paired and out of the E slice. Since this normally requires U moves, I end up doing a LOT of z2 moves. As I said, this isn't the best way to pair up edges, but it works. Note that if you end up with a parity case you are better off temporarily moving the edges to the M slice, shifting R and L in by an appropriate amount, and doing the parity normally. This is possible because you don't need to do single-layer R and L turns.

So I pair up edges in two steps on 6x6 and 7x7, three steps on 8x8 and 9x9, and four steps on 10x10. On the last step, I just use the standard 5x5 pairing (which is on M), because if I have the right and left grips on the far outside of the cube I can do slice moves AND R and L turns.

3x3

The 3x3 stage is pretty standard, press space and just solve it normally. On even cubes there are parity issues to deal with. The way I do this is, for OLL parity, shift L and R in until the r and l slices turn half of the cube, and then the algorithm is:
r U2 x r U2 r U2 r' U2 l U2 r' U2 r U2 r' U2 r'
and for PLL parity, just shift R in so that the r slice is halfway, and then do:
z r2 z' r2 U2 r2 y2 L2 y2 U2 r2 z r2 z'.
There are a lot of cube rotations so it looks longer than it is, but it's still kinda slow.

So, good luck with bigcubes! Make sure to give each one a try, because for a lot of them you'll get on the top 100 list if you solve it, no matter how long it takes.

Centers Last (old)

(Disclaimer: I wrote this guide in June 2007, on twistypuzzles.com. Here it is, with just a few edits. I haven't changed up the method much if at all since then.)

This is a method for solving arbitrarily large cubes. I decided to make a topic because I want to keep a permanent reference for this method, and because I just recently improved it a bit and I thought I'd like to share that with you.

From personal experience, it becomes faster than the reduction method for cubes of a certain size or greater, but that size is determined by a lot of factors (such as how fast you can make slice moves and how good you are at finding pieces for reduction). Suffice it to say, though, that if I ever attempt a truly huge cube (say 40x40x40) I'm not going to use reduction. This method does use about 30-50% more moves than a reduction strategy for a given size of cube, though, and although you have better recognition you will need more physical endurance to be able to complete a cube with this strategy.

The basic method is just this:
1) Solve all of the edges and corners. They must be solved relative to the centers on an odd cube.
2) Solve all of the centers without disturbing the edges and corners (much).

There are a lot of subtleties, though. Each step has many different ways to complete it, and there are a lot of small (but very useful) shortcuts. Throughout this guide, I'm going to use the r* notation; what this means is, on an NxNxN cube, to do a clockwise turn on all R-side slices from the second slice to the floor(N/2)th slice. For example, on a 6x6x6 cube r*' would mean to turn the 2nd and 3rd R slices counterclockwise. This is useful for generalized algorithms.

My current step 1 is done by matching up pieces of edge groups. I'll mostly be using 3x3x3-like moves for this part, so think of everything as corners, edge groups, and center groups. Using only the outer layers, move an edge group with a lot of similar pieces to DF (but if you can't find one in 4-5 seconds don't bother), and choose a color pair of edges that you will place in that spot. Make sure to remember which color is on top. Now, looking at the R and L layers, find edge pieces which match that color pair and move them to UF or UB using R/U/L moves, then shoot them down to DF with the appropriate slice move. Once DF is completely grouped up, you can do something like R' L F to move it into the R or L layer; keep doing this, while never disturbing the previously grouped edge groups. When you have 7 of them, you will not be able to do this any more; then, move the M layer to the E layer (just do z), and group another edge group together using horizontal slices. You can use moves like RUR' to get the fifth unfinished edge group into E. When you are done, move the finished group into U or D. At this point, U and D should each have 4 finished edge groups. You can actually finish one more edge group in E quickly (remember, you can do things like R2 because centers are irrelevant) before you have to do moves like (vertical slice) R F' U R' F (vertical slice)' to finish off the remaining edges. Note that you can only be sure of solving 11 of 12 edges like this. Don't try to use last-two-tredges or parity algorithms yet, though.

When you have paired up at least 11 edges, it's time to solve the edges and corners into a cage. Do it like the 3x3x3 step. Make sure it's solved relative to the centers if you're doing an odd cube. Here is a trick: when you get to the orient step (OLL for Fridrich, step 3 for Petrus) it's very helpful if the last edge group, the one which will almost certainly have parities in it, is one of the edges you are going to orient. If you're not color neutral, you can force this easily: for example, I solve Fridrich with a yellow last layer, so I can simply pair up all of the edge groups which don't contain yellow first, which will ensure that my parity edge is in the last layer when I get to it. So fix all of the orientation parities at the same time, then do OLL, then fix the permutation parity if you need to, then do PLL. The permutation parity is very easy, by the way: just do r*2 U2 r*2 U2 r*2.

So edges and corners are all solved; now onto step 2. In this step, we use commutators to solve all of the centers. The goal is to solve one center completely, then the opposite center, and then any other center, an adjacent center, and the last two. But again, there are a lot of subtleties and tricks here. First of all, a lot of people tend to use commutators like r U' r' d r U r' d'. There's nothing wrong with that, but I much prefer commutators in the form of r U l' U' r' U l and r U' l' U r' U' l. These commutators are very fast to execute because all of the interior slices are on the same axis, and you can see what is going on by only looking at two faces (where the centers are coming from and where they are going), leaving you free to look for the next center on an adjacent face. You'll have to experiment with this algorithm to get a feel for it, but you should know a couple things: (a) you can vary the affected centers in the 3-cycle by varying which slices you use (and the r- and l-like slices don't even have to be the same depth, i.e. r U 3r U' r' U 3r works fine); (b) x-centers can only be solved by one of these, but others can be solved by both; (c) the first move of the commutator should move a center piece that you want to move to the U layer into a space in U that does not already have a U-color center in it; (d) it's possible to bring entire blocks up to U by using more than one slice in place of one (or both) of the component slices of the commutator, but the first move has to bring the block to a place in U that is mostly or fully unoccupied, and some blocks are simply too large to do this with; and (e) it's best to only use setup moves involving U and only try to solve one piece or block at a time, so that the edges and corners don't get further than an AUF away from solved.

Now that we've gone over the small-scale details, let's concentrate on the large-scale ones. When you're solving the first center, pieces can come from all five centers, but you will notice that bringing them from the opposite center is time-consuming for a couple of reasons, notably 180-degree turns and slow recognition. There is a solution, however: the move r* F2 r*' l*' F2 l* will swap the F and D centers, completely on an even cube and mostly on an odd cube, although it messes up some edges so you will have to undo it after you solve the U center. So one way you could do the first center group (say white) is to bring every white center cubie from one face into the white face, then swap that face's center group (which contains no white cubies now) with the center group opposite white, and finally just bring all of the white centers (from the center group that used to be on bottom) onto the white face before reswapping the two center groups. Personally I like to hold the target face on U, and then bring all the target center pieces on F and R into the target face before doing y2 and doing it again. Once you have the first center, the opposite center is the same, but this time every center piece you want to bring in will be in an adjacent face, which makes it a little easier.

Now, you have four centers left to do. Pick one and hold it on U, then transfer the target pieces from F to U, swap F and D, transfer the target pieces from F to U, swap F and D, do y2, transfer the target pieces from F to U. If you want to be fast, this should be mostly mechanical, as it's a lot of the same type of move done from more or less the same angle. Now pick an adjacent face and hold it on U, with the just-finished face on B. Transfer the target pieces from F to U, swap F and D, transfer, swap. Remember that if you do one swap, you must do a second one, because otherwise the edges will get messed up. Now hold the last two centers on F and U (the order is irrelevant) and finish the cube. You're done!

By the way, if you want to use this on cubes smaller than 6x6x6, there are a couple of key differences. Depending on how good you are at reduction, it may be faster for you to just hold the edges on the E layer and pair them up normally. If that works for you, do it. Also, on the smaller cubes, you shouldn't bother with swapping centers, because there are not enough centers to swap to make it worthwhile. If you want to reduce the number of centers you have to move from one face to an opposite one, you can try to find a face that has as many centers as possible on that face and the adjacent ones, and start solving there. But unless you are very slow at the centers part of reduction, this will probably be slower for you for small cubes, so I don't recommend it for the small ones.

Solving 4x4x4 and 5x5x5 Supercubes

Although the 4x4 and 5x5 supercubes look much more difficult and impressive than the standard ones (even experienced bigcubers will wonder how you can find pieces so quickly!), they aren't much harder than the normal cubes: all they take is a slightly different method. The following tips and tricks were all gotten through experience and practice. Note that I'm going to describe the reduction method below, for a couple reasons: I think it's the fastest known way to do supercubes (and normal cubes for that matter), I'm most familiar with it, and you're probably most familiar with it. So if you're not a reduction user, then sorry, the only advice I can give you is to go practice :)

Centers

Centers are the only part that's really different about supercubes, although they are not the only part that you have to solve differently. In a supercube, every center has to be solved into the correct spot - they're not interchangeable anymore. Thus, instead of just solving blocks of centers, you have to actually pay attention to what goes where.

In a 4x4, you only have x-centers, which have three colors in the standard Pochmann sticker design (which I really recommend, because just drawing arrows or cutting corners off the stickers makes it very hard to see what goes where). I always put 4x4 centers together by making two 1x2 blocks. This probably doesn't require any explanation at all, although you might find it helpful to know that if you pair up the two yellow centers across a red-yellow 'edge', the other stickers will be paired across an orange-yellow 'edge' (since orange is the opposite of yellow). This can make recognition and setup easier.

If you go to solve centers, you might notice that the last center doesn't automatically solve itself anymore. That should happen 5/6 of the time, but fortunately there are algorithms to fix this quickly. If you want to swap Ufr and Ufl, you can use this A-perm-like algorithm:
x r' U r' d2 r U' r' d2 r2
If you want to do a diagonal swap, on the other hand, the algorithm I currently use comes from the Square-1 and is as follows:
r2 U D r2 U r2 U D r2 U r2 U D r2.
So your centers should be solved and you can now start on the edges step.

In a 5x5, on the other hand, you have many different types of centers, and now the fixed center, although it cannot move, must be oriented along with the rest of them. Again I pretty much solve the centers with my normal method (1x3 blocks, with the first block containing the fixed center) although it takes a bit longer to find the exact pieces I am looking for. If you see any blocks already formed for the first center or two, that will be very helpful, so try to preserve them.

On 5x5 you still have last center problems, but since PLLs are hard to recognize and turning a fixed center (only) by 180 degrees is slow, I prefer the following method. First I solve a 2x3 block on the second to last center, and position it so I can do rUr'-type triggers without messing it up. Then I intuitively solve all of the T-centers on U and F. It's important to solve them relative to the fixed center (not just relative to each other). There is one other case here, that of parity, where you have two edges swapped. The quick solution is just this:
r U2 r U2 r U2 r U2 r.
That algorithm should do an opposite swap of edges on U (a Usl-Usr swap to be exact), so now you can be certain to be able to solve the edges only. Finally, I have six T-centers left; I do an F turn to set them up to a more convenient location, and then finish off the centers with U turns and the Niklas commutator (which has two forms):
r U l' U' r' U l
l' U' r U l U' r'.
Sometimes it is actually more convenient to just solve the F centers with this commutator, and then finish off U with an A perm (which does exactly what you think it does). If you get an E or an H, good luck, you might as well just do two As. The A perms (for 5x5) are:
x r' U r' d2 r U' r' d2 r2
x r2 d2 r U r' d2 r U' r.

Edges

If you got the Eastsheen supercubes, you might notice that your edges and corners have more colors than they need to. The extra colors don't affect anything (although sometimes they can make recognition for PLL a little easier, on 3x3), and after a couple of dozen solves you'll learn to ignore them.

For 4x4, you can do edges exactly as you would on a standard cube. There is no change at all no matter what method you use, so go full speed on this step.

For 5x5, if you use AvG or any other kind of 2-pair edges method, do whatever you normally do. You'll notice that you never get parity - this is because we fixed it during the centers step. Convenient, isn't it? If the T-centers are solved, you can never get parity on the wings. Anyway, if you use the freeslice/bigcubes method, the biggest difference is that during the first 8 edges you can't do half turns of the faces (F2/R2/B2/L2 if you solve in the E slice, for instance) unless the slices are all solved. Just making them parallel isn't enough, because that will actually slightly mess up two centers. You will also notice that, during the last 4 edges, you can't do the m U2 m' U2 algorithm, because that also messes up centers a bit. (If you really want to do it, you should do the inverse algorithm twice.) You should also be careful during the last 2 edges: if you solve this in one step, be very careful about whether your algorithm messes up centers or not, because some will and some won't. If your algorithm does mess up centers, you can always just fix the edges with the r [flip edge] r' algorithm, which is completely center-safe.

3x3x3

The final step on both 4x4 and 5x5 is to solve a supercube 3x3. You don't need to learn any new tricks for this, you just have to be more careful. When you're putting together the cross, make sure that the centers are all oriented the right way; since this takes so much more thought than a normal cross I do not at all expect you to do this in one look. Feel free to do it one or two edges at a time. Now, if the centers on the first two layers are correctly oriented, you are basically home free: almost all F2L algs will keep them solved (the only one I know of that messes them up is the 2-gen edge insert algorithm, but you can always use an alternate algorithm), and most OLL algs will keep them solved as well. Some PLL algs do, and for certain ones (U perm, for instance) it might be helpful to learn alternate algorithms that don't mess up the centers at all, since you have to spend extra time fixing it. Here's one possible alternate U perm:
F2 U (M' U2 M U2 M' U2 M) U F2.
If you do end up with a messed up center, you should twist both it and the U center, because the U center hasn't been fixed to a specific orientation yet anyway. The following (intuitive) algorithm will flip the front center clockwise and the left one counterclockwise (note that here M and E mean every slice together):
M' U M E M' U' M E'.

There are three parity algorithms that you should know. The first two are the standard 4x4 parity algorithms (so of course they only show up on even supercubes). For the orientation parity, you can still use the normal one although that will twist the R or L center by a quarter turn, so I strongly suggest that you use some kind of slice-turn-only parity. My favorite is this one:
2R U2 x 2R U2 2R U2 2R' U2 2L U2 2L' x' U2 2R U2 2R' U2 2R'.
For PLL parity, the normal algorithm will mess up centers pretty badly, so here is one that won't:
x' 2U' R F' U R' F 2U 2D F' R U' F R' 2D' x
Finally, at the very end on both 4x4 and 5x5 supercubes, half of the time the U center will be twisted by 180 degrees. There are two algorithms I use for this, either of which can be faster depending on the cube and your style:
(R L U2 R' L' U)2
(R' U' R U')5.

And with that, you know everything you need to solve the 4x4 and 5x5 supercubes every time. Good luck, and good times!

Advanced Counting: Burnside's Lemma

Sometimes the theory we have discussed so far is not enough to calculate the number of positions of a puzzle. Fortunately there is a theorem from group theory that will help us with this. It is called Burnside's Lemma (although he was not the first to discover or prove it):

Theorem: Suppose we have a finite set X, and a finite group G of permutations of X. For each element g in G write |Fix(g)| for the number of elements of X that are not changed by g, and write |G| for the number of elements of G. Then the number of orbits of X is equal to 1/|G| Σ_g∈G |Fix(g)|.

What does this mean, though? Well, when we are counting the number of positions of a puzzle, we want the number of positions regardless of orientation - you shouldn't be able to get from one position to another just by rotating the puzzle, but we also don't want to miss any. The way we do this is by setting X to the set of every possible position in every possible orientation and G the possible ways to rotate the cube. The number of elements in X is relatively easy to calculate (it's like finding the number of positions of a puzzle, except that you don't fix anything; for example the size of X for the 2x2 cube would be 8! 3^7), and the number of elements in G can mostly be looked up from a small table, since it is 12 for a puzzle with tetrahedral symmetry, 24 for puzzle with the symmetry of a cube or octahedron, and 60 for one with the symmetry of a dodecahedron or icosahedron. Now, the number of orbits is simply the number of positions of a puzzle regardless of orientation, which is exactly what we want - so as long as we can compute the Fix(g) values, we know exactly how many positions a puzzle has, even if the puzzle has repeated pieces.

One really important thing to notice is that, for any cycle in a permutation in G, every piece on the cycle has to look the same. This is pretty important because it tells us that if we have, say, three of each piece, it is impossible for g to fix any positions if g contains a 4-cycle (or bigger), because there are no sets of 4 of the same piece. Also, if g contains only 3-cycles, and the number of duplicate pieces in each set is not a multiple of 3, g cannot fix any positions.

I want to do two examples to show how this works. First I'll do the very simple example of a cube with three red faces and three blue faces which can be switched any way we want. We already saw that our basic approximation gave something like .83 when the actual answer was 2. Let's use Burnside's Lemma. The size of X is 6 choose 3 or 20, and the size of G is 24 because the puzzle is cubical. Let's calculate the |Fix(g)|'s: if g is the permutation that does nothing, 20 positions remain the same; if g is a rotation around a corner (there are 8 of these), 2 positions are fixed; if g is a 90-degree rotation around a face (there are 6 of these), there are no fixed positions; if g is a 180-degree rotation around a face (there are 3 of these), there are 4 fixed positions; and if g is a flip around an edge (there are 6 of these), there are no fixed positions. So the total number of positions is 1/24 (20 + 8*2 + 0 + 3*4 + 0) = 1/24 (48) = 2. It's exact!

Now let's do a very hard example: the (non-jumbled) Little Chop or 24-cube, which has 24 pieces, with 4 each of 6 colors and no orientation. The size of X is 24!/(4!^6) and the size of G is 24 again. Calculate the values of |Fix(g)|: if g does nothing, 24!/(4!^6) positions remain the same; if g rotates around a corner, we create 8 3-cycles so no positions can be fixed; if g is a 90-degree rotation around a face, there are 6 4-cycles so 6! positions are fixed; if g is a 180-degree rotation around a face, there are 12 2-cycles so 12!/(2!^6) positions are fixed; and if g is a flip around an edge, there are again 12 2-cycles so 12!/(2!^6) positions are fixed. The total number of positions is 1/24 (24!/(4!^6) + 6*6! + 3*12!/(2!^6) + 6*12!/(2!^6)), which works out to 135,277,941,853,080. We can't really verify this, but if we have done our calculations right it is the correct value. Note that with our basic computations, fixing one of the 4 pieces of a specific color, we would get an approximate value of 1/4 (23!/(3! * 4!^5)) = 135,277,939,046,250, which is pretty close (but not exact).