Bas Verseveldt
Member
Note that this is not a very accurate approximation as you do need to memorize misoriented pieces, and you do need to execute algorithms to solve them, hence they affect both memorization and execution time. Parity will only affect execution time (but as the chance of parity is 1/2, we can easily deal with it).i asked this 6 years ago but i didnt get an answer, ive been looking for it and the answer seems to be 10.xx, but i cant trust myself
whats the average number of targets if you are using 2 fixed buffers? (dont mind orientations; passive pieces unoriented count as 0 targets, parity edge and corner count as 1 each and breaking into a new n cycle counts as n+1 targets)
also whats the average of passive pieces to twist? not counting the buffers i guess, or whatever is easier to calculate
Anyway, I brute forced this problem by varying the number of pieces P from 3 to 10 (unfortunately my computer couldn't handle 12 due to there being many permutations) and got a very clear linear trend T = 1.0916P - 1.7731 for the average number of targets T.
For P = 8 (corners) the exact average is 6.968 while the model predicts 6.959 targets, so it is clearly very accurate.
For P = 12 (3x3 edges) this model gives an estimated 11.32 targets with an estimated error smaller than 0.02. So it is definitely not 10.xx.
For P = 24 (big cube wings) the estimate is 24.4 targets (altough I'm not sure how accurate that estimate is as it is extrapolated so far)
This formula does not apply for big cube centers, because there are identical pieces which this formula does not take into account. For those, I would use the approximation that it is possible to avoid cycle breaks altogether (which is practically always the case) and that we cannot choose our orientation (like on a 5x5). In that approximation, the average number of solved pieces can be computed exactly. Every piece has a chance of 1/6 to be solved, so an average of 23 * 1/6 = 3.83 pieces is solved (remember: not 24, because we never don't consider the buffer to be solved). Neglecting cycle breaks this gives us an average of 23 - 3.83 = 19.17 targets.
On a 4x4 we can choose the orientation based on centers, and I don't know a nice way of handling the problem then, as we cannot consider the rotations to be independent positions. From experience, I guess that the average number of targets is about 15 if we choose the optimal orientation.