[Help Thread] Big Cubes Discussion

[goodatthis]

I'll just do 10 and 7....go ahead and guess something in between for 8:30.

For 10: 1:30/4:00/2:30/1:00/:50 is about 9:50

For 7: 1:15/2:15/2:00/:45/:45 is about 7 minutes.

I currently average 5 and my splits are 1/1:15/1:30/:30/:30

Thanks! By the way, that mini shengshou you sold me really helped me get into 7x7 more, I just couldn't get faster with a full size, so thanks!

 

[Christopher Mowla]

I can't make heads or tails of that page. The diagrams don't make sense to me. I have no idea what cubies are trying to be solved, there is no "before and after" and it is just not intuitive for someone at my level.

For Wing Edges
​

Have you tried my wing edge last layer method for the wing edges of the nxnxn? (Attached to the end of this post.) I provide images like Thom's but also permutation diagrams. (I use 5x5x5 images for the cases as well.)

My instructions for "Step 1A" which says to solve two wing edges simultaneously can be ignored...you can just use one of the algorithms listed to solve a single wing edge (if one isn't solved already) and then go to "Step 1B" to solve the second edge in that tredge/composite edge of the 5x5x5.

That is, this is a 3 step/algorithm method, but you can simply make it into a 3-4 step method. Unless noted other wise, all algorithms in my guide work on the nxnxn.

This is the only written last layer wing edge method which tells you how to visually solve every case, apart from "Step 1A", which you can ignore its instructions and just use the above instructions to solve some cases easier using 4 steps instead of 3.

For Middle Edges​

I show in steps 6 and 7 of my commutator 3x3x3 solution how to solve all cases (assuming the corners are solved already).

There is a great page for 4x4x4 algos that is very easy to understand:

http://www.speedsolving.com/wiki/index.php/4x4x4_Parity_Algorithms

Is there an equivalent page for 5x5x5?

There isn't currently, although I don't see why there shouldn't be. I can simply just copy the 4x4x4 parity algorithms page, create 5x5x5 last layer images, convert all algorithms to the 5x5x5 (and recount their moves if in the process of them to the 5x5x5 increase move count, which does happen for some algorithms), omit all algorithms which work only on the 4x4x4, and omit sections which only apply to the 4x4x4. This page would only be for the wing edges of the 5x5x5, as the middle edges are not considered parity cases on big odd cubes.

 

[uberCuber]

I can't make heads or tails of that page. The diagrams don't make sense to me. I have no idea what cubies are trying to be solved, there is no "before and after" and it is just not intuitive for someone at my level.

The "before" is the picture shown, the "after" is a fully solved cube. Any cubies that are not solved in the picture are the ones the alg will solve. Only pay attention to the "3-cycles" section right now, the first section of algs labeled "First/Second edge tricks" isn't something you should worry about. In the 3-cycle section it should be clear which 3 cubies aren't solved in the picture, so those are the 3 pieces that the alg cycles around without messing up the rest of the cube at all.

I did r R U R' U' r' R' U' R U and somehow this did not do the trick. I undid it and tried the other formulas and they only made things much worse, so my guess is I am not doing the inverse operation properly.

TDM is correct. Think about what an inverse function means: taking a solved cube and doing R U R' U' followed by R' U' R U doesn't result in a solved cube, so it wouldn't make much sense for that to be the inverse, right? To take the inverse of an alg, reverse the order of moves along with changing clockwise to counterclockwise and vice-versa. The inverse of (R U F' L2 D R') would be (R D' L2 F U' R').

So which way is "M" and which way is M' when you move all 3 slices?

I guess that M was the moving of the top towards the front and M' was the front towards the top.

Whether "M" is referring to 3 slices or just one doesn't affect the direction. Yes, your description is correct.

In a second I'll link you to a full alg.cubing.net animation of the solution I described earlier so you can see what was intended, but first a quick word on notation to avoid more confusion:
In my earlier post, and in the K4 page that I linked, the convention was that "Rw is a 2-layer wide turn" and "r is a 1-layer inner slice turn".
alg.cubing.net uses SiGN notation, where "r is a 2-layer wide turn" and "2R is a 1-layer inner slice turn".
I am still meaning "M" to mean all 3 middle layers.

Here is the animation, hopefully this thing works

Spoiler: A bit of explanation about edge 3-cycles

Like I mentioned in my other post, the goal is for solutions like the one I linked above to be intuitive, so here's some stuff to guide your thinking in the right direction.

In this discussion I will use the SiGN notation I mentioned above, so things like 2R are single inner slice turns.

Lets examine probably the simplest cycle: the first commutator I used in the above solution, [2R, R U R' U'].
Take a solved cube (doesn't matter if 4x4 or 5x5), and start with yellow top and green front so that I can refer to specific colors to make this easier to understand. Perform [R U R' U', 2R] = R U R' U' 2R U R U' R' 2R' to set up the case. (The inverse of a commutator [A, B] is the commutator [B, A].) You should see 3 edge pieces cycled around, where the blue edge on front needs to go to the back, the orange on back needs to go to the right, and the green on right needs to go front.

Now for the commutator that can solve this case. Notice that two of the 3 pieces to be cycled, the blue and orange pieces, are sitting on the same slice (the 2R slice to be exact), while the third, green piece is not on that same slice. This is great: to make an easy edge 3-cycle, you need a single "interchange" move that swaps two of the pieces without disturbing the third, and you need a few-move "insertion" that moves the third piece into the spot of one of the first two without disturbing the rest of the "interchange" layer. For this case, 2R will be our interchange move, and R U R' U' will be our insertion. Notice that if you do R U R' U', the effect of replacing the back edge with the one on the right is the *only* effect had on the 2R layer; every other piece on the 2R layer remains unchanged. This is a necessary quality of any insertion you choose, take note of this for later.

The A and B of our commutator will be the interchange and insertion, 2R and R U R' U'. But you have to think to see which should come first. The insertion is a setup to the back position, so think about things from the perspective of that back position: the interchange moves the front piece into that back position. Before you do the insertion, the interchange is moving a piece into "the position that should be blue." Notice that after you do the insertion, the interchange is moving a piece into "the position that should be orange." In this case, since the piece on front is blue, you need to do the interchange first. So the solution is [2R, R U R' U']:

2R - solve the blue piece into its position, and move the orange piece out of the way
R U R' U' - replace the back spot that the interchange is affecting with "the position that should be orange"
2R' - undo interchange to solve the orange piece into its position
U R U' R' - undo insertion to restore cube

You can take the commutator we just looked at and change it a little bit to affect a different few pieces:
[2L', R U R' U']
[2L', L' U' L U]
[2L, R' U' R U]
etc.

Sometimes there is no interchange between two pieces that doesn't disturb the third, but you can do a setup move to make it so there is one:
[B2: [2R2, B R B']] = B2 2R2 B R B' 2R2 B R' B' B2

Well, there you go. Past this, experiment on your own, try to figure out your own solutions to other 3-cycles pictured on the K4 page, study the given solutions if you can't come up with something. If something in that explanation of how that simple commutator works doesn't make sense, let me know. If you understood that but have trouble understanding a different case, keep thinking and trying different stuff before asking here.

When you are trying to solve all the edges in an actual solve, start out by just picking one piece, using a commutator to solve that piece into the right spot, then picking another piece, and so on. Make sure that each commutator you use doesn't mess up pieces you have already solved unless you are doing it on purpose.

You are correct about M and M', but again, i think people sometimes use M to mean -just- the middle slice in this context.

This time I actually specified in the post that I wanted all 3 slices for M, that way it solved the full blue tredge rather than just the middle piece. However that alg is also useful with just the single middle slice as it then cycles 3 midges. I'm not actually sure what the notation is supposed to be for all 3 slices vs just the single middle one...

 

[maps600]

Wait......Your Edge pairing is incomplete...
You did you f3l without completing your edges...
so just destory your f3l and pair up these by use the flip algorithm,just like the 4x4.....you slice an inner layer flip the pieces and slice back
and then you might end up in a parity error ... use the same alg and solve that case and then solve like a 3x3
Hope this helped

You should be knowing the flip alg.....
R U R' Sledge hammer

He's not trying to use Reduction.

 

[unsolved]

The inverse of R U R' U' is U R U' R', not R' U' R U. You invert the entire sequence of moves, not each move individually.

Yes, thank you. When I went back and re-read this post:

http://www.speedsolving.com/forum/showthread.php?34858-Help-thread-Big-Cube-Discussion&p=1001036&viewfull=1#post1001036

...I realized the error of my ways.

When you are doing r, are you moving both the inner right slice and the outer right slice? I think r in this context at least means to turn the inner right slice only.

I have always rotated just the inner slice when seeing a move such as "r" and the outer face when I see "R".

You are correct about M and M', but again, i think people sometimes use M to mean -just- the middle slice in this context.

I thought the post indicated a 3x3x3 "perm" algo was being applied to the 5x5x5 holistically, meaning you treat all 3 inner slices as just the one middle edge on the 3x3x3.

Could the group weigh in on this and provide clarification?

The edges of a 4x4x4 will correspond to the inner and outer wing edges on your 5x5x5. Hope this helps )

Brilliant! I did not make that connection, THAT is what I was missing!

So, for the Last Layer on the 5x5x5, is it best to solve the middle edges, then deal with the wings as a 5x5x5?

Is there a list of perms for 5x5x5 middle edges somewhere, you know, like that awesome 4x4x4 reference:

http://www.speedsolving.com/wiki/index.php/4x4x4_Parity_Algorithms


Many thanks to those who have helped me out. I am determined to evolve some way to solve this 5x5x5 thing!

 

[uberCuber]

I thought the post indicated a 3x3x3 "perm" algo was being applied to the 5x5x5 holistically, meaning you treat all 3 inner slices as just the one middle edge on the 3x3x3.

Could the group weigh in on this and provide clarification?

Yes, as I clarified at the bottom of the last post I made (did you see that post?), I had meant for all 3 slices to be moved together because all 3 blue edges were already attached to each other and could be solved as one 3x3x3 edge. However if you only wanted to move the middle edge without moving the wings, you can apply that same alg using only the middle-most slice instead of all 3 slices. Many <M,U> algs let you do make this decision on 5x5, allowing some extra control over which pieces you move around.

So, for the Last Layer on the 5x5x5, is it best to solve the middle edges, then deal with the wings as a 5x5x5?

Probably yes. (though I think you mean deal with the wings like a 4x4x4?)

Is there a list of perms for 5x5x5 middle edges somewhere, you know, like that awesome 4x4x4 reference:

You don't need much. The middle edges can be treated as 3x3 edges, which means they aren't as complicated to solve as the wings are. They could all be solved in 1 alg with 3x3x3 ELL (I think it's 27 algs), but if you don't feel like learning all of that, here's some useful stuff you can use to do it in 2 looks. Note that for all the algs with M moves here, it doesn't matter if you do 1 slice or all 3, so if one option would result in some extra wings being solved, go with that.

some 3-cycles:
[R U R' U', M'] and its inverse and mirror
M' U' M U2 M' U' M and its inverse
U perms from 3x3x3

Also these algs that just flip 2 edges:
adjacent edges (UR - UB): R U R' U' Rw' U2 R U R U' R2 U2 Rw (where Rw is turning all layers but the left-most one)
opposite edges (UF - UB): M' U M' U M' U2 M U M U M U2

 

[unsolved]

Here is the animation, hopefully this thing works

I wanted to comment on this first. I have 3 things to say:

1. Wow!
2. Awesome!
3. Thank you!

Those new features with the pause and the axis changes and annotation are really sweet! Who designed that page? And is there a set of instructions on how to encode an amazing replay like that?

(bows in Wayne's World fashion and says "We're not worthy.... we're not worthy....")



As for the rest of the post, I did not want to cite it in a long quote here to avoid the cumbersomeness. I spent a great deal of time going over the information, and it was an entire education unto itself. I still struggle with the [A,B] and [A:B] notation, and when a sequence has nested versions of them, I must write the entire sequence out or else I am boggled infinitely. After a good 90 minutes of trying, re-trying, and taking copious notes, I have made some serious strides.

Thank you for taking so much time to break it all down to a noob like me. That was very nice of you and I greatly appreciate it.

Did you not see my post response to you?

I did Chris, but that was back when I had no idea how to read the notation. I am revisiting the PDF file(s) now that the fog is starting to clear.

 

[unsolved]

As for centers, I really only have a couple of algos for moving things around.

To exchange center Top-6 with center Front-6:
r U' M' U r' U' M U (please check to see if I have the M and M' correct, I still confuse which way to spin them sometimes)

To exchange center Top-4 with center Front-4:
l' U' M' U l U' M U

I know how to exchange the center-8 from the top and bottom faces:
f2 D (the middle slice from top to right)2 D' f2 D (the middle slice from top to right)2 D'

The centers from 1-3-7-9 behave like center moves on the 4x4x4 so I did not have to learn anything new there.

Does anyone else know any cool center algos that do not disturb anything else on the 5x5x5?

For example, I can only solve centers 2 and 8 from top to front by rotating them as needed to fit the above.

Is there a way to do the center exchange of top-2 o front-2? Or top-8 to front-8?

 

[uberCuber]

For example, I can only solve centers 2 and 8 from top to front by rotating them as needed to fit the above.

Is there a way to do the center exchange of top-2 o front-2? Or top-8 to front-8?

(yes your M directions are correct)

The center algs you listed are also 3-cycle commutators of the same form I was discussing for edges earlier. It's just that here, two of the three pieces being cycled are the same color, so it ends up looking like only two pieces were swapped. For instance, the first alg you listed, r U' M' U r' U' M U, does the 3-cycle [Top-6 -> Front-2 -> Front-6].

Notice that that alg can be written as the commutator [r, U' M' U]. Using the terminology from my edges discussion, the "r" move is the interchange (two of the pieces in that 3-cycle are on the r slice), and the U' M' U is the insertion (which inserts the Front-2 piece into the Top-6 spot).

Anyway, all you need to do to get different swaps is take that commutator, [r, U' M' U], and do any number of these things to it:
1. Switch the first slice move to a different slice
2. Switch the second slice move to a different slice
3. Switch the direction of the U moves (e.g. U' M' U becomes U M' U')

Try out different combinations and get a feel for what's going on. See what combination(s) result in those 2-2 or 8-8 swaps you wanted. See which changes have no apparent effect on "which 2 are swapped," and then try to think if the full 3-cycle that is being done might be different even though the apparent 2-swap is the same.

And for more fun, have the commutator use all 3 slices instead of just 2. i.e. Have the interchange turn one slice and the insertion turn both of the other slices simultaneously.

 

[unsolved]

The center algs you listed are also 3-cycle commutators of the same form I was discussing for edges earlier. It's just that here, two of the three pieces being cycled are the same color, so it ends up looking like only two pieces were swapped. For instance, the first alg you listed, r U' M' U r' U' M U, does the 3-cycle [Top-6 -> Front-2 -> Front-6].

Wow, I would have never noticed that. I guess on even larger cubes you can't get away with "cheats" such as this.

Anyway, all you need to do to get different swaps is take that commutator, [r, U' M' U], and do any number of these things to it:
1. Switch the first slice move to a different slice
2. Switch the second slice move to a different slice
3. Switch the direction of the U moves (e.g. U' M' U becomes U M' U')

Try out different combinations and get a feel for what's going on.

I will give it a shot. But when I experiment, weird things happen. Take a look at this:

I combed the 4x4x4 parity algos pages, cmowla's 3-cycles.pdf, and his 5x5x5 PDF, and I saw nothing that could help out with this. This can't be that uncommon for the remaining edges on a 5x5x5 that nobody has a solution.

Ugh! Any ideas?

 

[Genesis]

Wow, I would have never noticed that. I guess on even larger cubes you can't get away with "cheats" such as this.



I will give it a shot. But when I experiment, weird things happen. Take a look at this:

I combed the 4x4x4 parity algos pages, cmowla's 3-cycles.pdf, and his 5x5x5 PDF, and I saw nothing that could help out with this. This can't be that uncommon for the remaining edges on a 5x5x5 that nobody has a solution.

Ugh! Any ideas?

Solve one of the edge first, then the other using parity algs(Depending on your alg for 4x4, it may or may not work)

 

[unsolved]

Solve one of the edge first, then the other using parity algs(Depending on your alg for 4x4, it may or may not work)

There is no 2-cycle for the case I have shown.

 

[Genesis]

There is no 2-cycle for the case I have shown.

Solve one off the edge, ignoring what the other turns out to be
E.g. Set up such that you can solve the yellow green edge, then undo it, leaving you with only the yellow orange edge

 

[qaz]

you need to flip the two middle edges and swap the two wings.

to flip the two edges (M is one layer)
R' U R //setup flipped edges on UR/UL
(M' U)4 (U M)4 //flip
R' U' R // undo setup

to swap the wings:
R B //setup to wings on the right of UF/UB tredges
l' U2 l' U2 F2 l' F2 r U2 r' U2 l2 //swap
B' R' //undo setup

 

[Christopher Mowla]

Wow, I would have never noticed that. I guess on even larger cubes you can't get away with "cheats" such as this.



I will give it a shot. But when I experiment, weird things happen. Take a look at this:

Spoiler

I combed the 4x4x4 parity algos pages, cmowla's 3-cycles.pdf, and his 5x5x5 PDF, and I saw nothing that could help out with this. This can't be that uncommon for the remaining edges on a 5x5x5 that nobody has a solution.

Ugh! Any ideas?

I'm not sure what you mean.

I assume you are calling this:

For Wing Edges

Have you tried my wing edge last layer method for the wing edges of the nxnxn? (Attached to the end of this post.) I provide images like Thom's but also permutation diagrams. (I use 5x5x5 images for the cases as well.)

my 5x5x5 PDF.

If you look at the top right corner of page 12, two algorithms for this exact case are listed.

All you have to do is add a y cube rotation before and after the algorithm(s) (there are two listed). For example, with the first algorithm (in Old WCA notation):
y'
x' r' U R U r' U' R' U' B2 r' B2 l U2 l' U2 r2 x
y

(My method handles all 8! possible wing edge positions.)

And I am pretty sure that I explain in my 3x3x3 commutator solution

For Middle Edges
I show in steps 6 and 7 of my commutator 3x3x3 solution how to solve all cases (assuming the corners are solved already).

that you can orient two adjacent middle edges using two Niklas commutators in the following manner:
y2 x' [M', U L' U'] z2 [M', U' R U] z2 x y2 (in Old WCA Notation)

Combine both of these two together to solve/generate your specific case:
y2 x' [M', U L' U'] z2 [M', U' R U] z2 x y2
y'
x' r' U R U r' U' R' U' B2 r' B2 l U2 l' U2 r2 x
y

Ed, if you like, you can post some example 5x5x5 last layer wing edge scrambles (or images of cases as you have been), and I can make a YouTube video specifically for you to explain how to use my PDF guide to solve those cases.

 

[unsolved]

I'm not sure what you mean.
If you look at the top right corner of page 12, two algorithms for this exact case are listed.

All you have to do is add a y cube rotation before and after the algorithm(s) (there are two listed). For example, with the first algorithm (in Old WCA notation):
y'
x' r' U R U r' U' R' U' B2 r' B2 l U2 l' U2 r2 x
y

If this is the case, I just didn't recognize the rotated version in the source document. I always solved the 2 middle edges that needed flipping with:

M F R F' M' F R F' M' B' R B M B' R' B

...but again, I am not sure if I have my M and M' notation correct. I am easily confused, repeatedly, when trying to remember the "clockwise" direction of a slice that has no such direction since it is in the middle.

Ed, if you like, you can post some example 5x5x5 last layer wing edge scrambles (or images of cases as you have been), and I can make a YouTube video specifically for you to explain how to use my PDF guide to solve those cases.

I appreciate the offer Chris, but I really need to buckle down and learn how to use what is already out there! This is now the second time I thought I encountered an undocumented instance of something that is a rotated or mirrored version of something that already has been taken care of.

It is kind of incredible that this one position would require 32 moves to solve. 20% shy of twice God's Number for the 3x3x3!

 

[uberCuber]

M follows L in the alphabet, much like "clockwise" M follows the direction of clockwise L

 

[unsolved]

Ed, if you like, you can post some example 5x5x5 last layer wing edge scrambles (or images of cases as you have been), and I can make a YouTube video specifically for you to explain how to use my PDF guide to solve those cases.

Chris,

If that offer still stands, I'd like to see a solution to this without any "rotating the cube" notation. Somehow I keep messing up my middle edge flips so I must be doing something wrong.

I am getting closer!

 

[tseitsei]

Chris,

If that offer still stands, I'd like to see a solution to this without any "rotating the cube" notation. Somehow I keep messing up my middle edge flips so I must be doing something wrong.

[images here]

I am getting closer!

I would do:

Holding the cube blue in front and yellow up:

(note that here M and M' means only the middle slice, NOT the 3 innermost slices...)
(R B) M' U M' U M' U2 M U M U M U2 (B' R') that solves midges

(and here r and l mean right and left inner slices only)
U' r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2 U that should solve the rest of it...

 

[Christopher Mowla]

Chris,

If that offer still stands, I'd like to see a solution to this without any "rotating the cube" notation. Somehow I keep messing up my middle edge flips so I must be doing something wrong.

Spoiler

I am getting closer!

This is probably one of the shortest combo algorithms for this case.
(2R2 B2 U2 2L U2 2R' U2 2R U2 F2 2R F2 2L' B2 2R2)(R2 U B' R 3D R' B U' R B' 3D' B R)

If you're wondering from a theoretical perspective how long of algorithms we (I) are capable of making by hand which are a single algorithm to solve the same case, I just found (in about 45 minutes) an algorithm of the same length as the combo above in the slice half turn metric.
F2 D' 2B2 3F U2 2B U2 2F' U2 2B' R2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F U' 3F' 2B2 D F2

However, using the base of this algorithm, I create an algorithm which is 23 single slice half turns, for a case which the minimum combo is (a 15 move adjacent 2-cycle algorithm) + (the 13 move adjacent middle edge flip algorithm) = 28 moves.
2-3b2 U 2B U2 2F' U2 2B' R2 U' 2F' 3R2 U2 3R U' R2 U 3R' U2 3R2 2F 2-3b2

For an even permutation example, below is an algorithm to a 3-cycle of wing edges and a two adjacent middle edge flip which is 4 moves shorter than (a 10 move 3-cycle of wings) + (a 12 move 2 opposite middle edge flip)
R' [B' F2 3R' F2 3R2 2D B, U2] R

This tells us that algorithms which solves middle edges and wing edges simultaneously is unpredictable as far as comparing move count.