Back when I was 12 years old I got obsessed with solving the Rubik’s Cube and eventually figured out a method for it. Normally these sorts of solutions are just sort of bad, but because I hadn’t figured out commutators as a concept there aren’t any inverse sequences done and everything is ‘forward’, resulting in a very unusual and hopefully interesting approach.
The ground rules I was playing with were that I didn’t want any external help whatsoever, which for some reason included using a pen and paper to make notes. I felt I’d been given a bit of a spoiler by being told about layer by layer. Because of this the method doesn’t so much give sequences of things to do but notes as to what to do in different situations which feed into each other nicely. Most of these were found by trial and error. Anyhow, here’s the instructions:
Solve the bottom cross. This is done intuitively as in most solutions.
Solve the bottom corners. Again this is done intuitively.
Solve the middle edges. This is done with the sequence FU2F’U2L’U’L which puts the FU edge into FL and its mirror image F’U2FU2RUR’ which puts the FU edge into FR. To use them position an edge which needs to go in the middle so that its top sticker matches the front center sticker and then use the sequence starting with F if it needs to go in the FL position or F’ if it needs to go in the FR position. If none of the edges on the top go in the middle but there are still edges in the middle positioned wrong use one of those sequences to put one of the edges on the top into the position of one of the misplaced edges in the middle and proceed from there.
Get the top edges positioned and oriented. The basic move for this is to use one of the sequences in step 3 to move an edge from the top layer to the middle layer then use the the other handed sequence to fix it. Starting this with an F’ is a ‘left break’ and starting it with an F is a ‘right break’. Each case is defined by a first move to do then you forget about what you just did and go back to the beginning of this step with a new observation of where everything is. These cases feed into each other in a way which is guaranteed to always eventually solve the top edges. The different cases are as follows:
If none of the top edges are in the correct orientation do a left break.
If all of the top edges are in the correct orientation rotate the top until either all four are in correct positions or exactly two are in correct positions. If all of them are positioned correctly you’re ready for the next step, otherwise
If the two incorrectly positioned edges are opposite each other position the cube so that one of the correctly oriented but wrongly positioned edges is in the front and do a left break.
Otherwise the two wrongly positioned edges will be next to each other. Position the cube as a whole such that one of them is in the front and the other is on the left and do a left break.
If exactly two of the top edges are correctly oriented and they’re adjacent to each other then rotate the top as few units as possible so that at least one of the edges is in the correct position, (in the case of at least one of the correctly oriented top edges already being in the right position this means don’t rotate the top at all). Then position the cube such that the remaining incorrectly positioned edge is in the front. If there is no such edge orient the cube so that either of the correctly oriented edges is in the front. Then do a break in the directly of the other correctly oriented edge.
If exactly two of the top edges are correctly oriented and they’re opposite each other then if neither of those edges is correctly oriented do a U2 to get at least one of them correctly positioned. If even that wouldn’t correctly position either of them then rotate the top until at least one of them is correctly positioned. Then:
If both of those edges are in the correct position orient the cube as a whole so that one of them is in the front and do a left break.
Otherwise orient the cube as a whole so that the correctly oriented but incorrectly positioned top edge is in the front and do a break in the direction where color of the sticker on the top edge matches the color of the front center sticker.
Position the top corners. To cycle UFR → UBR → UBL do a left break and start over. To cycle UFL → UBL → UBR do a right break and start over. It’s always possible to position the corners using these sequences at most twice.
Orient the top corners. The rotation sequence used for this turns UFR clockwise and UBR counterclockwise. It’s done by doing a left break and starting over. (This kicks off a cycle done in step 5 which you’ll then fix doing step 5 again.) To use this to solve the orientations in the fewest steps:
If all four of the top corners are oriented wrong do the rotation sequence so only two or three of the top corners are oriented wrong.
If three of the top corners are oriented wrong do the rotation sequence so that only two adjacent corners are oriented wrong.
If two non-adjacent corners are oriented wrong do the rotation sequence so that two adjacent corners are oriented wrong.
If only two adjacent corners are oriented wrong then if it’s possible to finish solving with a single use of the rotation move do that. Otherwise redesignate the side face they’re both adjacent to as the top and you’ll be able finish the solve with a single use of the rotation move.