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This results is quite intriguing, but "expected" i guess. However it will have no implications on how scrambling (or solving) will be done now or in the future.

Still pretty cool to have done fewest moves in less than 20 moves, altough the actual acramble of course had shorter optimal solution. One of my 19 solutions had a 18-turn optimal solution

This results is quite intriguing, but "expected" i guess. However it will have no implications on how scrambling (or solving) will be done now or in the future.

Still pretty cool to have done fewest moves in less than 20 moves, altough the actual acramble of course had shorter optimal solution. One of my 19 solutions had a 18-turn optimal solution

I'm happy to announce that our 15f* result (which accounted for a fair percentage of our overall computation) was just confirmed today by Thomas Scheunemann. This is a independent code, an independent team, and independent hardware. While it does not confirm our proof of 20 in its entirety, it does confirm a result that requires about 13% of our overall effort and about 80% of our codebase, including our set cover, our entire search phase, the prepass, and much more.

If there were an error in any single one of the coset runs (up to level 15), or if we omitted one or counted one twice, or if there was a problem in the set covering, these two results would not have agreed.

Awesome! Of course, by showing that the number of 21f* is zero, and getting all the way up to 15f*, you've made me wonder... how much more calculation would it take to get the last five numbers? I know the average is 17.7 or so, so most cubes will be in that range, and it will probably take many times what you've already done. Would it be closer to a month on a Google farm? A year? A decade?

Long term I plan to make this a BOINC project, and let whoever wants to work on it with their machines (and a nice graphical display of cubes per second or rotating, changing cubes or something like that).

To finish off the distribution, I've got some ideas up my sleeve---but I estimate it would probably take about 500 times the effort. So call it, for better or worse, 17 CPU millennium. The good news is Moore's law is getting us closer all the time.

The good news is we can find a lot of intermediate interesting things, like 16f*, 17f*, and indeed just enumerating the 20f*'s is interesting at some level. And just the slow and steady progress; every CPU night (say ten hours) grinds through some hundred billion positions or so (plus all the rotations of said positions) and will probably find a handful of new 20f*'s.

It may take a few years to get the project up and working fast enough (this is just a hobby after all), but who knows what will happen.

But maybe in five years, desktops cease to exist and everything is done on mobile platforms like phones and (relatively slow) tablets, and then BOINC goes nowhere.

Why does it have to be called "god's" algorithm... everyone knows its the FSM that is the higher power of the universe...
how long is the solution to the evil edge flip if the flipped all but for say two.. or flipped two corners...?

The new paper is in preparation. In the meantime, the technique is the same as was used for 22, with some technological improvements and one additional insight that helped. The paper for 22 is linked from the cube20 website, but here's the link again:

Been playing around a little with the "hardest scramble":
F U' F2 D' B U R' F' L D' R' U' L U B' D2 R' F U2 D2

What do I see?
U D edges flipped in place
middel layer edges flipped in "opposite positions"
Corners not quite, but after D2 U2:
All corners flipped in place
All edges "opposite flipped"

From the original scramble:

Simple route to cross: R L F B R L D U
U gives me two crosses!
Middle layer edges in position but flipped!
R and L can be OLL'd in f-SM-SM-f' style:
B L D L' D' L D L' D' B'
F R D R' D' R D R' D' F

Leaves a solved belt plus two solved faces plus 4 L-sides
L-sides have two pairs which can be alternately "solved" by L2 or R2
Bonus: you break an existing pair and build a new one
This looks so nice and I would like to believe there is an elegant finish to this cube pattern.

Anyway, back to OLL for R and L layers.
Who can tell me how to do this in one alg?
Note that these two sides seperately are actually "unsolvable": on both faces two edges need to be exchanged.
So this opposite-side-OLL would have to tackle this too...

What I did in my actual solve: set up - Z or H perm - undo set up to switch two edges on each face
Leaving both R and L face in combinations of N, V or F perm to finish the cube

Another question, how could I extract this -double-OLL-with-parity alg from Cube explorer?
I managed to get a 13 move finish after the D + U cross but how do I enter a "target" state?
I tried leaving grey facelets for all but the belt and the orange / red stickers to be OLL'ed but it would not accept it.

13 move finish after the double cross: L2 B2 L D2 L2 D2 B2 U2 F2 R2 F2 R' U2

it was suspected to be 20 but mearly solving all cases still isnt satisfying... a nice mathematical explanation would do and the guy/girl who does it will probably get a group theory prize too

Can't you just do all the different 1 move algs, then all the 2 move algs, then all the three move algs, etc. until you reach 43 quintillion, and if you own a computer, it shouldn't be too hard, so why did it take some google supercomputers to work it out?