Lars Vandenbergh's CubeZoneSpeedcubing taken one step further

This short article describes the results of a computer analysis I did for solving the cross of the first layer in the least number of moves. In this study we are trying to determine the required number of moves to the solve the cross for all possible cases if we would always be able to see an optimal solution (God's algorithm). From that information we can then calculate the average and maximum number of moves that a "perfect" cross solver would need to solve the cross from a random state.
Most crossfirst speedcubers always start with the same color when building the cross. A lot of people can still comfortably build the cross on the opposite face of their prefered color, and a few people are completely colorneutral and start building the cross on whichever face it seems the easiest. In this computer study we have investigated all three scenarios.
When solving the cross on the same face, there are 4 edge pieces that we need to take into account. They can each be oriented in 2 ways and can placed in 12 locations. The amount of cases we need to explore in this scenario is 2^{4} x (12 x 11 x 10 x 9) = 190,080. In the following table you can see how many cases can optimally be solved in a certain number of moves, both in face turn metric and quarter turn metric:
Face turn metric  Quarter turn metric  



Average: 5.81 moves  Average: 6.59 moves 
When solving the cross on either of two opposite faces, there are 8 edge pieces that we need to take into account. They can each be oriented in 2 ways and can placed in 12 locations. The amount of cases we need to explore in this scenario is 2^{8} x (12 x 11 x 10 x 9 x 8 x 7 x 6 x 5) = 5,109,350,400. In the following table you can see how many cases can optimally be solved in a certain number of moves, both in face turn metric and quarter turn metric:
Face turn metric  Quarter turn metric  



Average: 5.39 moves  Average: 6.15 moves 
When solving the cross on any of the six faces, there are 12 edge pieces that we need to take into account. They can each be oriented in 2 ways and can placed in 12 locations. However since we are using all the edge pieces and edge pieces can only be flipped in pairs, the orientation of the 12th edge can be derived from the orientation all the others. The amount of cases we need to explore in this scenario is 2^{11} x 12! = 980,995,276,800. In the following table you can see how many cases can optimally be solved in a certain number of moves, both in face turn metric and quarter turn metric:
Face turn metric  Quarter turn metric  



Average: 4.81 moves  Average: 5.50 moves 
This page is maintained by Lars Vandenbergh  Last update on 15th February 2015 