I have a 4x4 Rubik's cube (aka Rubik's Revenge), which I can solve by reducing it to a 3x3 and then solving the 3x3. However, this can result in "parity" cases, where a single edge of the 3x3 is flipped. To resolve these, I have to remember a complicated and unintuitive algorithm, which I find unsatisfying.

Because of this, I would like to learn a different method for solving the 4x4, which doesn't rely on reducing it to a 3x3. I know that such methods exist, but I haven't had much luck finding any good explanations of them. All of the explanations I can find are for variants of the reduction method. (E.g. the cage method, Yau method, etc. all involve pairing up the edges and then solving as a 3x3.)

Note that speed solving isn't really my thing. My goal here is to be able to forget about the cube completely and still be able to solve it in several years' time. (Perhaps with a bit of work, but without looking anything up.) I can solve the 3x3 this way, just by knowing a bunch of fairly intuitive algorithms that are enough to cover all cases. I'm looking for a similar set of algorithms for the 4x4.

(I did try reducing it to a 2x2 instead of a 3x3, which presumably would avoid the parity cases if I could get it to work. However, I didn't have much luck with this, as I can't assemble the last few blocks of the 2x2 without messing up the already completed ones. If someone has a way to do this it would be a nice solution.)

  • $\begingroup$ Has a useful answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio May 23 '18 at 8:22
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    $\begingroup$ @Rubio accepting an answer discourages further answers. The one given is quite useful, but I am hopeful that a more comprehensive one will come along also. Thus I choose not to accept it just yet. Thank you. $\endgroup$ – Nathaniel May 23 '18 at 8:31
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    $\begingroup$ Fair enough; just note that after the first week a question is posted, absent any other inducement for people to look at it again (e.g. a bounty), it's increasingly unlikely you're going to get any extra answers. If the existing answer is in some way lacking, perhaps give that feedback to the answerer, and/or update the question to highlight what you're ideally looking for in a more comprehensive answer. $\endgroup$ – Rubio May 23 '18 at 8:34
  • $\begingroup$ I dunno, I've had good answers to years old answers on other sites. But it's not that important in the end. $\endgroup$ – Nathaniel May 23 '18 at 8:51

Here are a couple interesting Youtube videos on solving the larger cubes layer by layer.

13x13x13 Solve Part 1: Layers 1 through 3 - by SuperAntoniovivaldi
Tutorial #5: How to Solve a 5x5x5 Layer by Layer (Part 1) - by europeancubers

However, no matter which way you try to solve the 4x4, you will run into parities of some form. For the classical single edge parity, a method I learnt of solving it is to move the second layer from the top into a 90 degree offset, and move around all of the centre pieces and edge pieces around on that layer so that the layer still stays at that rotational offset, yet the pieces are in the correct position. Doing so will remove the parity.

The best way of attempting something like that is to use some general cube solving techniques, such as commutators and conjugation

  • $\begingroup$ Perhaps I don't understand parity as well as I thought I did, then. I thought it was because when we pair two edges up to make a "dedge", we don't know whether we're putting them the right way round or not, because they look the same. So I figured if I just solve the corners first and put each edge directly in its correct place then I can easily avoid that situation (at the cost of needing longer algorithms), because if I put them the wrong way around they won't match up to the corners. Is that not correct? $\endgroup$ – Nathaniel May 13 '18 at 7:34
  • $\begingroup$ @Nathaniel That's indeed correct. If you use the Cage Method (corners -> edges -> centers) you won't run into parities. The algorithms are mostly commutators, though. And I'm not sure, but perhaps you might still have some kind of parities when you solve the last few edges/corners. But because you won't have to save the centers, those parity-algorithms might be a lot shorter/easier. $\endgroup$ – Kevin Cruijssen May 13 '18 at 8:08
  • $\begingroup$ if you try that method, you may still run into a parity. In order to solve the last two edges correctly, you would have to apply an algorithm that leaves an odd number of middle slices rotated 90 degrees. $\endgroup$ – micsthepick May 13 '18 at 8:11

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