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Wetten, dass..? vom 5. November
A while back I wanted to swap the yellow and blue face of my 4x4 because I didn't like its color scheme. I could've just peeled off the 2*16 stickers and rearrange them, but that's not necessary. Actually only 10 stickers need to be peeled off, the rest can be done by twisting the puzzle. The following algorithms, applied to a solved cube, make this obvious by rearranging most of the stickers and only leaving a few stickers on the wrong faces.
| Stefan's SwapAdjacentColors4x4 | ||
|---|---|---|
| 16 - 16 | Stefan Pochmann | |
| I found this using ACube, treating the cube as a Rubik's Domino. Almost only uses U and R turns! | ||
| Per's SwapAdjacentColors4x4 | ||
| 18 - 18 | Per Kristen Fredlund | |
| Per improved an algorithm of Grant Tregay to arrive at this cool algorithm. Read more. | ||
| Per's SwapOppositeColors4x4 | ||
| 10 - 7 | Per Kristen Fredlund | |
| Just in case you want to swap opposite faces, after this algorithm you only need to peel off and swap 8 stickers (instead of 32). Read more. | ||
| Per's SwapOppositeColors5x5 | ||
| 16 - 14 | Per Kristen Fredlund | |
| After this algorithm you only need to peel off and swap 10 stickers (instead of 50). Read more. | ||
| Stefan's SwapAdjacentColors5x5 | ||
| 79 - 64 | Stefan Pochmann | |
| After this algorithm you only need to peel off and swap 15 stickers (instead of 50). Read more. | ||
Stefan Pochmann
Last modified: March 20 2007, 21:28:10