Hello, fellow speedsolvers! My name is David Smith, and I posted an introduction earlier today. It's a pleasure to meet you!
I have discovered a result for counting the permutations of the Rubik's Cube which I hope might be of some interest to the community. It is based on the work of Chris Hardwick, who I previously collaborated with regarding the n x n x n cube formulas he found, as well as T. R. Keane and H. J. Kamack, who wrote a paper called "The Rubik Tesseract". I hope that posting this here is relevant and of value to the speedsolving community; if not I apologize in advance.
Basically, I have discovered a general formula which counts the number of visually distinguishable reachable positions of a Rubik's Cube of any size and any dimension. It turns out that Rubik's Cubes can be defined in four dimensions and higher. Many of you may be aware of a computer program called Magic Cube 4D, which is a fully functional representation of a 4-dimensional Rubik's Cube. There is also a program called MagicCube5d, which is the analogous simulation of a 5-dimensional Rubik's Cube. In fact, some have even solved these 5D cubes! The largest one to be solved so far is the 6^5, and its solution is over 1 million moves long.
Theoretically, a Rubik's Cube of any dimension can exist, and these higher dimensional cubes can be of different sizes just like the 3D cube, so one could have for example a 4^6 cube (a 6-dimensional cube with 4 pieces per edge, like the Rubik's Revenge). Just as Chris Hardwick thought to generalize the well-known permutation counts of the 3x3x3, 4x4x4, and 5x5x5 to cubes of arbitrary size, I thought that perhaps I could do the same with cubes of higher dimensions.
I started with the problem of finding the number of permutations of a 3x3x3x3 Rubik's Cube, and then generalized this to an n^4 Rubik's Cube. This formula can be seen here:
http://www.gravitation3d.com/david/n^4_Cube.pdf
As you can see, I liked Chris's use of the (n mod 2) term in his formulas, which equals zero when n is even, and one when n is odd, as a way of accounting for different cases in one formula. Thanks, Chris!
I then later found a formula for n^5 cubes:
http://www.gravitation3d.com/david/n^5_Cube.pdf
and n^6 cubes:
http://www.gravitation3d.com/david/n^6_Cube.pdf
After much work, I finally found a general formula for counting the number of permutations of an n^d cube, with n >= 2 and d >= 3.
http://www.gravitation3d.com/david/n^d_Cube.pdf
I should definitely mention that my results are only upper bounds, as the only way to prove them exact would be to somehow find general algorithms for cubes of any size which would demonstrate that the possible positions implied by my formulas are all reachable, or possibly by making heavy use of group theory, in which I am much a novice. In the past these algorithms have always been found, once the impossible positions have been removed, for many different types of 3D and 4D puzzles, including the n^3 cube, the Megaminx, and Magic120Cell, a 4D version of the Megaminx.
I apologize for the lack of a proof for any of these formulas (although I have proved them for myself), they would certainly be long and detailed. Actually, I do have a paper on the derivation of the n^4 cube formula, which can be found here. All of these results and others can be found on my website, www.mathproofs.bravehost.com. By the way, some of you may find some of the puzzles on the puzzles page interesting. Many of them seem completely impossible to solve, especially the last one.
Describing how each formula works would be a bit complicated (and very lengthy!), but I would be more than happy to answer any questions you may have. Also, feel free to raise any concerns or doubts you may have about my formulas, I completely welcome constructive criticism!
Thanks for reading this long post, I hope my work is a positive contribution to the community, albeit an abstract one. I would also like to thank Chris in particular for his help all those years ago; without finding his formulas I may never have had the idea to come up with my own.
I have discovered a result for counting the permutations of the Rubik's Cube which I hope might be of some interest to the community. It is based on the work of Chris Hardwick, who I previously collaborated with regarding the n x n x n cube formulas he found, as well as T. R. Keane and H. J. Kamack, who wrote a paper called "The Rubik Tesseract". I hope that posting this here is relevant and of value to the speedsolving community; if not I apologize in advance.
Basically, I have discovered a general formula which counts the number of visually distinguishable reachable positions of a Rubik's Cube of any size and any dimension. It turns out that Rubik's Cubes can be defined in four dimensions and higher. Many of you may be aware of a computer program called Magic Cube 4D, which is a fully functional representation of a 4-dimensional Rubik's Cube. There is also a program called MagicCube5d, which is the analogous simulation of a 5-dimensional Rubik's Cube. In fact, some have even solved these 5D cubes! The largest one to be solved so far is the 6^5, and its solution is over 1 million moves long.
Theoretically, a Rubik's Cube of any dimension can exist, and these higher dimensional cubes can be of different sizes just like the 3D cube, so one could have for example a 4^6 cube (a 6-dimensional cube with 4 pieces per edge, like the Rubik's Revenge). Just as Chris Hardwick thought to generalize the well-known permutation counts of the 3x3x3, 4x4x4, and 5x5x5 to cubes of arbitrary size, I thought that perhaps I could do the same with cubes of higher dimensions.
I started with the problem of finding the number of permutations of a 3x3x3x3 Rubik's Cube, and then generalized this to an n^4 Rubik's Cube. This formula can be seen here:
http://www.gravitation3d.com/david/n^4_Cube.pdf
As you can see, I liked Chris's use of the (n mod 2) term in his formulas, which equals zero when n is even, and one when n is odd, as a way of accounting for different cases in one formula. Thanks, Chris!
I then later found a formula for n^5 cubes:
http://www.gravitation3d.com/david/n^5_Cube.pdf
and n^6 cubes:
http://www.gravitation3d.com/david/n^6_Cube.pdf
After much work, I finally found a general formula for counting the number of permutations of an n^d cube, with n >= 2 and d >= 3.
http://www.gravitation3d.com/david/n^d_Cube.pdf
I should definitely mention that my results are only upper bounds, as the only way to prove them exact would be to somehow find general algorithms for cubes of any size which would demonstrate that the possible positions implied by my formulas are all reachable, or possibly by making heavy use of group theory, in which I am much a novice. In the past these algorithms have always been found, once the impossible positions have been removed, for many different types of 3D and 4D puzzles, including the n^3 cube, the Megaminx, and Magic120Cell, a 4D version of the Megaminx.
I apologize for the lack of a proof for any of these formulas (although I have proved them for myself), they would certainly be long and detailed. Actually, I do have a paper on the derivation of the n^4 cube formula, which can be found here. All of these results and others can be found on my website, www.mathproofs.bravehost.com. By the way, some of you may find some of the puzzles on the puzzles page interesting. Many of them seem completely impossible to solve, especially the last one.
Describing how each formula works would be a bit complicated (and very lengthy!), but I would be more than happy to answer any questions you may have. Also, feel free to raise any concerns or doubts you may have about my formulas, I completely welcome constructive criticism!
Thanks for reading this long post, I hope my work is a positive contribution to the community, albeit an abstract one. I would also like to thank Chris in particular for his help all those years ago; without finding his formulas I may never have had the idea to come up with my own.
