Kirjava's 3x3x5 Method

[Kirjava]

So the 'standard' method seems to be to solve the inners and middle then solve U and D as a domino. It's about time I explained the method that me and Dan Cohen have been getting sub30 with.

The steps are 3x3x3 -> Columns -> L4E

The first step is to solve it as a 3x3x3. This allows all layers to be turned freely and just leaves the inner slices to solve. It's probably quite similar to the other method only you're solving the outer layers instead of the inner ones. You can use any method here, but some have delicious advantages that are 3x3x5 specific :3. I hear that F2L recognition is a bit annoying.

After this step is done, you want to orient your cube so that the 'top' and 'bottom' faces are on the L and R sides so you can use r and l moves for the inner layers.

Next step is columns (usually). The aim is to solve the edges and centres (the two types of inner layer pieces) in the first two layers leaving four edges left to solve. This is accomplished with a technique called columns.

The idea of the technique is to blockbuild the pieces together. Generally, I'll first solve one of the two 1x2x3 'blocks' then move onto the other. This is usually done by creating a 1x2x2 centre-edge-centre block and then adding the centre-edge pair. This leaves you with another block to do the same with on the other side.

The second block is a little harder. The first 1x2x2 goes in just fine, but the final centre-edge pair presents an interesting problem. It cannot be easily created as the other columns have been because you are restricted in where you can store pieces (all the empty space is in use). From here there are a few things you can do. At first I was just placing the edge and then using [TPerm, l] to comm the centre in. These days it mostly varies on the case. You can put the edge in and use some variation on r U2 r U2 r2 U2 r2 U2 r U2 r' to solve the centre or put the centre in and use some commutator to solve the edge. (With the latter technique you might get lucky and get the chance to force a L4E skip.)

I forgot to mention that the final pair may be easy to create. When that happens it feels like you're skipping a step or something.

When you get good, you can treat the last step as last 5 edges and three centres - looking for the best way to solve from there instead of moving to L4E. (However, L4E is often the best solution anyway.)

Finally, here's an alg for every L4E case;

3 Cycle
[l, U' R U x U R2 U' x']
or l' U2 r' D2 r U2 r' D2 r l

2x2 Cycle Diag (PLL Parity)
R2 F2 U2 r2 U2 F2 R2

2x2 Cycle Adj (2 Flip)
U2 r' U2 l r U2 r' U2 r U2 l' r' U2 r

2x2 Cycle Opp (PLL Parity + 2 Flip)
l U2 r D2 r' U2 l' r U2 l D2 l' U2 r'

2 Cycle Diag
r U2 r' U2 r' U2 l U2 r' U2 r U2 F2 r2 F2 l'

2 Cycle Adj
r' U2 x l U2 l' U2 x' r2 U2 r U2 r' U2 F2 r2 F2

2 Cycle Opp
r U2 r U2 x U2 r U2 l' x' U2 l U2 r2

4 Cycle Clockwise
r U2 r2 U2 r' U2 r U2 r' U2 r2 U2 r

4 Cycle Anticlockwise
r' U2 r2 U2 r U2 r' U2 r U2 r2 U2 r'

4 Cycle X (British Parity)
r2 B2 r' U2 r' U2 B2 r' B2 r B2 r' B2 r2 B2

4 Cycle Z
r U2 r l' U2 r2 U2 r' U2 r U2 r' U2 r' U2 l U2 r'

The badasses that know K4LL should already know most of these. You'll have to mirror/invert them to cover everything.

[qqwref]

Cool method, and I'm glad you wrote it up. I notice that your "columns" thing is essentially the old domino-Roux method, which goes left 1x2x3 -> right 1x2x3 -> "CLL" (last four corners). Do you ever get parity (inners solved, outers an R2 off) using this solution? Or does the AUF before L4E take care of it?

Have you considered solving the last center while solving *any* edge? You'd still be left with four edges (or sometimes three) although there would be slightly different/more cases. Alternatively, you could just blindly solve the last edge and then do the last five edges in whatever way is easiest. There are some relatively nice 3-cycles (such as U2 l' D2 l U2 l' D2 l) so although there are 120 possible cases most of them could be done without using too many moves.

[Kirjava]

Cool method, and I'm glad you wrote it up. I notice that your "columns" thing is essentially the old domino-Roux method, which goes left 1x2x3 -> right 1x2x3 -> "CLL" (last four corners).

Ja, this is what I did when I first solved a domino.

Do you ever get parity (inners solved, outers an R2 off) using this solution? Or does the AUF before L4E take care of it?

You mean U2? :P I can only imagine that would happen with this case; R2 F2 U2 r2 U2 F2 R2. I don't see why you wouldn't just do a U2 then solve it. AUF pretty much takes care of it O_o.

Have you considered solving the last center while solving *any* edge? You'd still be left with four edges (or sometimes three) although there would be slightly different/more cases. Alternatively, you could just blindly solve the last edge and then do the last five edges in whatever way is easiest. There are some relatively nice 3-cycles (such as U2 l' D2 l U2 l' D2 l) so although there are 120 possible cases most of them could be done without using too many moves.

Sure, I touched on this in the post. I do either of these from time to time. The method I use is really just 1x2x3 -> 1x2x2 -> 5 edges 3 centres. L4E is the fallback. The problem with doing something like this though is that it uses up thinking time which gets expensive when you're getting down to 20 seconds, so it's really only worth it if you can see that there will be three edges left to solve or something else special.

[maggot]

ive been using this method, but i hadnt derived the algs for the L4E. . i typically just intuitively use 4x4 edge parity and single edge swap. i tried using coming up with a few variations of last 2 edge groups on 5x5, but i was having difficulty trying to modify the algs. thanks for the algs for the L4E, i will learn them.

and after you solve the 3x3, do a z move and make a "cross" on any color, you can solve any 4 of the slots with r U2 r' U2 r U2 r' U2...r U2 r' U2 r U2... over and over until you make a pair and then insert it into its corresponding slot. or, use the mirror. problem is, there is much more efficient ways to make the blocks on the 1st 3 'pairs'. r U2 r' U2... will solve the cases, but you can solve in much less moves if you use a little brainpower.

thanks again, this is one of my more favored puzzles.

[Athefre]

I was solving this same way when I had mine.

[Kirjava]

and after you solve the 3x3, do a z move and make a "cross" on any color, you can solve any 4 of the slots with r U2 r' U2 r U2 r' U2...r U2 r' U2 r U2... over and over until you make a pair and then insert it into its corresponding slot. or, use the mirror. problem is, there is much more efficient ways to make the blocks on the 1st 3 'pairs'. r U2 r' U2... will solve the cases, but you can solve in much less moves if you use a little brainpower.

I was going to suggest placing two centres first as a 'beginner' method for the second step, but that's a bad habit to get into.

I was solving this same way when I had mine.

I imagine many people have found something similar to this. However, I doubt they have developed it to this extent. This guide was requested by a few people after they saw my videos.

[rahulkadukar]

@Kirjava : How would you handle this case ? I think I am getting it because my cube might have popped.



Only the 2 adjacent middle layer pieces need to be swapped

[TMOY]

This case is unsolvable, sorry. You need to disassemble your cube.