Dear Mods: this may be a non-research question, but I am asking it because from my knowledge, the question appears nontrivial. This video claims that a Rubik's cube can be solved from any starting position with the repeated application of two fixed moves. The method is easily seen as wrong because the specific moves he suggests actually leave the bottom right $2\times 2 \times 3$ block unaffected. My question is more general: does there exist a finite set of moves which if applied repeatedly from any starting position, yield a solved Rubik's cube?Moves can be thought of as permutations on the set configurations of a Rubik's cube. A solved Rubik's cube is a specific configuration of interest. If these moves get you from any configuration to a solved configuration, then (by symmetry) these moves must get you from any configuration to any other configuration. This means these moves constitute a permutation of maximum period. Does such a permutation even exist? What is known about the existence of such a permutation?
Permutations of maximum period: applications to Rubik's cube
co.combinatorics;board games
Copy/paste from another (slightly different) cstheory question on Rubik's cube:What is the shortest sequence that scans all cube's configurations, or does the configuration graph of the cube have a Hamiltonian path? is a special case of a more general unsolved problem: For an arbitrary permutation group with a given generating set, is its Cayley graph Hamiltonian?There are some good info available online, you can download the lecture notes by W.D.Joyner: Mathematics of the Rubik's cube (downloadable in pdf format).Chapter 6 of the lecture notes is devoted to Cayley graphs and God's algorithm.EDIT: If I correctly interpret the question Does such a permutation even exist?, the answer is yes (the Cayley graph of a finite group is connected); a trivial argument:let $k$ be the total number of wrong configurations;for every wrong configuration $W_i, 1 \leq i \leq k$ write down the solving sequence of moves $S_i$;let $S_i^{-1}$ be the inverse sequence of $S_i$;obviously the (long) sequence $S_1 \cdot S_1^{-1} \cdot S_2 \cdot S_2^{-1} ... \cdot S_{k-1}^{-1} \cdot S_{k-1}^{-1} \cdot S_k$ will lead you to the solved configuration from any wrong starting configuration, but you must look at the cube to see when to stop (there is no synchronizing sequence).