Random Cubing Discussion

What's the number of possible LS positions, disregarding pre-AUF? I can't figure out how to not double count AUF psotions, I always get confused

if the corner is in U, always bring it above the last slot

for cases where the corner is in place, it can be in one of 3 orientations, and the edge can be in place oriented, in place flipped, in U oriented, or in U flipped. One of these cases is solved so 3*4 - 1 = 11 cases with the corner in place

for cases where the corner is in U, it can still be in one of 3 orientations, and the edge can be in place oriented, in place flipped, or in any of the 4 U positions flipped or oriented. 3*10 = 30 cases with the corner in U

11 + 30 = 41
basically all these: http://algdb.net/Set/F2L
 
if the corner is in U, always bring it above the last slot

for cases where the corner is in place, it can be in one of 3 orientations, and the edge can be in place oriented, in place flipped, in U oriented, or in U flipped. One of these cases is solved so 3*4 - 1 = 11 cases with the corner in place

for cases where the corner is in U, it can still be in one of 3 orientations, and the edge can be in place oriented, in place flipped, or in any of the 4 U positions flipped or oriented. 3*10 = 30 cases with the corner in U

11 + 30 = 41
basically all these: http://algdb.net/Set/F2L

Doh...I meant including LL CO and EO
 
Doh...I meant including LL CO and EO

so full OLS instead of LS?

you can use AUF for the cases with both LS pieces permuted, which there are 5 of

edge flipped, corner solved: put the only flipped or only oriented edge in front, then there are 3^3 = 27 CO cases, 54 total cases
edge flipped, corner oriented cw: put only flipped or only oriented edge in front, then still 3^3 = 27 CO cases, still 54 total
edge flipped, corner oriented ccw: yeah same thing again so 2 * 3^3 = 54 total cases
edge solved, corner oriented cw: 8 CO cases not including AUF, but now the edge cases are L, line, dot, and cross. Normally this would be 108 cases but you can use symmetries to reduce this. For cross cases, this is just a subset of CLS, which has 8 cases. Dot also has 8 cases. L has no rotational symmetry so it's just 27 cases. For line cases, place the CO first, and then you'll have either a vertical or horizontal line. However, for the CO case with 90 degree rotational symmetry, you only have 1 case, so there are 7*2 + 1 = 15 cases. The total is 8 + 8 + 15 + 27 = 58 cases
edge solved, corner oriented ccw: This is just the mirror of the last case so also 58.

The total of these is 54*3 + 58*2 = 278

For cases where you cannot AUF, it's a lot easier because it's just 8 EO cases and 27 CO cases for all 36 of them, which makes 36*8*27 = 7776


So the total for OLS is 8054
 
The maximum possible TPS is around 947,045,759 TPS.
Maximum "radius" of a cube is diagonal to center, so 57 cm * sqrt 2 * 0.5 * 0.01 m/cm = 0.403050865 m
Multiply by pi/2 for length of arc: 0.633110819 m
divide by speed of light: about 2.1 ns
1/x: 947,045,759 TPS
Edit: oops unit fail multiply answer by 10:
9,470,457,590 TPS

Was this somehow related to the following quote

Seriously no one got this idea?? (Everything is hypothetical)(cube it friction less)
Assume that the corners (of the corners of the cube) are moving at light speed and the cube is regular 57mm.the corners would be (57/2)*sqrt2 away form the axis of rotation.therefore the circumference would be (57/2)*sqrt2*2pi = 57*sqrt*pi.dividing it by 4 to get the length traveled by the corner in one Quater turn (57*sqrt2*pi/4).Dividing this by speed of light (in mm/s) to get the minimum time taken for one quarter.(2.112 x 10^-10 sec per move)And then multiplying this value with god number ie. 26 (in QTM). The final answer is 5.491 * 10^-9 seconds or approx 5.5 nano seconds
 
Did OLL parity and lost my grip and damaged the sticker in a rather interesting way...


image.jpg
 
So I have these transparent Aolong V2 force cubes and I'm curious, how much do you guys think they're worth? I'm thinking about selling them on the Buy, Sell, Trade forum eventually since I really only use the blue one. They aren't stickered but are all lubed and tensioned and mostly unused, only turned to see if the tensioning was good.
 
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