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The first time I tried solving the Fisher Cube, I encountered an interesting problem on the last layer. This was something I never came across on the Windmill Cube and in retrospect, I have no clue how I fixed it. I think I messed up the whole lot and didn’t have it the second time through. In searching online for parity fixes, most people find themselves in a very different position. This suggests their solution is very different from mine. The good news is that I don’t get the other parity error using my method. This should be the only parity alg I need to worry about. I’ve tossed it here as I’m going to forget this. Anyways, here it is…I found it on a German YouTube video.

(R U’2 R’ U)*2 (z) L’ U’ L

This entry was posted on June 24, 2012 at 12:38 and is filed under Algorithms, Blog, Cubing with tags Fisher Cube. You can follow any responses to this entry through the RSS 2.0 feed You can leave a response, or trackback from your own site.