Solving should always happen in quadratic time. The number of pieces scale as n^2 after all. I guess it's fair to assume that memorization is quadratic in terms of number of elements memorized, so quartic in this case, but this is probably harder to check.
This is interesting. For pieces you can say that there are
6n^2 - 12n + 8 pieces to the n x n x n cube. Of those we expect the following number of pieces to be solved from the start (law of probabilities).
Of the edges or wings there are: floor[(n-2)/2] wing orbits. Each wing orbit is expected to have 1 piece solved after the scramble, and I mean mathematical expectation here.
Each center orbit is expected to have 4 pieces solved (this is using mathematical expectation, but using an approximate method that Daniel Beyer and I worked out when studying BH probabilities).
There are floor[(n-2)^2/4] center orbits, so this gives 4*floor[(n-2)^2/4] solved centers.
If you have an odd cube then you can expect one of the centralmost edges to be solved half the time (the mathematical expectation of pieces being in the solved location is 1, but it is flipped half the time). This gives 0.5 expected solved central edges. There are 1/3 solved corners expected by the same reasoning.
Let's assume n is even first to get rid of the cumbersome notation.
I am examining the following equation:
[Number of pieces] - [expected number of solved pieces] = [number of pieces to be solved during a solve]
6n^2 -12n + 8 - [(n-2)/2 + 4*[(n-2)^2/4] + 1/3] = (30n^2 - 51n +28)/6 if n is even
--edit--
Just realized that this approach is wrong on the even case, since you can rotate the cube to maximize solved pieces. Ignore the even case, the odd case is really the one to go by. I'm a bit lost as to how to approach the even case now.
--edit--
If n is odd you get
[Number of pieces] - [expected number of solved pieces] = [number of pieces to be solved during a solve]
6n^2 -12n + 8 - [(n-3)/2 + 4*[((n-2)^2-1)/4] + 1/3 + 1/2] = (30n^2 - 51n +34)/6 if n is odd
Again this formula is approximate only because I don't know how to exactly calculate the number of expected solved centers. I only estimate it to be that 1/6 of each orbit is solved (every piece has an approximately 1/6 chance to land in a "solved" location if placed randomly on the cube).
These formulas count the mathematically expected number of unsolved pieces on the cube. These pieces must be memorized and solved. I was simply interested in knowing more precisely how increasing the size of the cube increases the amount of memorization and solving that must be done.
Chris
Seriously - why are you so slow at 5x5x5 execution? You're a way faster cuber than me (even at 5x5x5) - I can't imagine why you'd be so slow at 5x5x5 execution.
. And usually i don't have any huge recall delays. I'll film myself the next time i try a 5x5x5 bld to see where i'm losing so much time.
