3 ms·
| As the cube gets larger, the complexity will get reduced, ( by ratio of volume/surface area ) ? Am I right ? I know nothing about this except what a Rubik's
by blastrat 10y ago
| As the cube gets larger, the complexity will get reduced, ( by ratio of volume/surface area ) ? Am I right ?
I know nothing about this except what a Rubik's cube looks like, but at first cut I'd say your comment is wrong but on the right track.
Rubik's cubes don't depend on volume at all, but the ratio of increasing the surface area is still an inverse square, so you'd be right if you focused on the ratio between area to linear dimension rather than the ratio of volume to area...
- gregfjohnson 10y agoBy "as the cube gets larger" do you have in mind 4x4x4 .. NxNxN cubes? The largest (physical) cube I have according to that metric is a 7x7x7 cube. I have the following: 1x1x1 ;-) 2x2x2 3x3x3 4x4x4 5x5x5 7x7x7 Going from 3x3x3 to 4x4x4, there was one additional algebraically distinct sub-puzzle that I needed to solve. Above 4x4x4, those same algorithms solve the 5x5x5 and 7x7x7. I have a marvelous proof (but don't have room in this margin) that these algorithms will work for all larger cubes. (Just kidding..) I'm sure there are increasing numbers of moves that do multiple things simultaneously as N gets larger, but my approach is to ploddingly move 3 cubies at a time leaving the rest of the cube unchanged, and then around the corners and edges flipping two cubies at a time. (The new move I needed for the 4x4x4 was to simply rotate one of the middle slices to switch the parity of the edges if it was wrong.)
- jozydapozy 10y agoThe 4x4x4 is far more difficult then the 3x3x3. The 5x5x5 is almost the same as the 3x3x3 with just one extra step, though it takes more time to solve. The cubes with even numbers (4x4x4, 6x6x6) are more difficult because these have no fixed centers.