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[Help Thread] Solving 5x5 LBL using commutators, how would I figure out what commutator to use?


Nov 8, 2013
Hi, I’m back after a very long hiatus from the forum. I’m thinking of posting pictures of my cube collection

So, I’m very familiar with the reduction method, it’s my default for 4x4 and up. But I thought of solving my 5x5, which turns much more smoothly than my smaller cubes, using a LBL(layer by layer) method. I could figure out the first layer intuitively, but now, I’m in the sea of commutators that is the middle 3 layers. I figured that I would be doing something like this for each layer:

  1. Center pieces of the layer if I don’t have it already
  2. Inner edge pieces of the layer
  3. Outer edge pieces of the layer
I did the first layer in such a way that I already have the center of the second layer on each of the 4 sides, but I know that I won’t always be in the position of already having the center of the layer done. I’m trying to get the inner edge pieces of the second layer in place, but I can’t seem to figure out which commutators to do or if I need conjugates. I understand the concept of commutators and conjugates, but can’t figure out how to execute it basically. I think I need more than just in depth videos to figure this out. Here’s my 5x5 as it is now with the first layer solved intuitively:

Last edited:

Christopher Mowla

Premium Member
Sep 17, 2009
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Conjugated commutators are probably the easiest approach to doing anything like this. You can simply use variants and conjugations of Niklas to handle any case. I don't know if you have seen my video Easy Beginner Tutorial on Rubik's Cube Commutators, but it may help you to get the general "motion of moves" concept in place. (The majority of that video is about solving the 3x3x3 with commutators, but I do jump to the 4x4x4 a few times to explain how the concept is the same.)

But TLDR, you can use this commutator. R U' 2R U R' U' 2R' U. And a variant to affect the top edges for later. R U' 2L' U R' U' 2L U.

I also created a complete algorithm set to solve both wings with one algorithm in 13 moves or less. (Only one case requires 13 moves. The reset are 12 or fewer.) But solving "layer by layer" naturally doesn't include this type of task. (You clearly want to solve all bottom wing edges, then the midges, then the top wings.)