• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 35,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

God's numbers for partial solves

Joined
Oct 3, 2008
Messages
2,068
Likes
6
Location
Toronto (Canada)
WCA
2009HOLT01
YouTube
riffz
#21
Mmm yes. I would be interested in hearing move distributions for colour neutral first layers compared to ones using pseudoblocks, but I'm really only interested in the half turn away scenario.
 
Joined
Oct 8, 2006
Messages
914
Likes
15
Location
Malden, MA, USA
WCA
2006NORS01
YouTube
cuBerBruce
#22
To check my (in)efficiency at FMC I would like to know God’s number for the following sub steps:
- F2L- slot with no edges oriented, to “leaving 3 corners”
- F2L- slot with two edges oriented, to “leaving 3 corners”
- F2L- slot with all edges oriented, to “leaving 3 corners”

Same for “leaving 5 corners”

To keep things “simple” let’s assume the remaining corners should not be twisted in place.
I've previously run JAcube on all 69 non-trivial cases (mod pre-AUF, post-AUF, and reflections) of solving the edges after F2L minus one slot. I find that whether you're interested in no edges misoriented, 2 edges misoriented, or 4 edges misoriented, there exist cases that require 9 moves to solve just the edges. After counting the pre-AUF and post-AUF, possibly 11 moves might be required. I have verified, however, that for no edges misoriented, that solving the edges requires at most 10 moves. (I've used the assumption that a last layer edge in the slot is always considered oriented.)

Existing software doesn't readily permit directly calculating for situations like leaving a corner 3-cycle or leaving a corner 5-cycle. One approach may be to search for solutions for solving the edges, and check resulting corner configuration for each solution.
 

AvGalen

Premium Member
Joined
Jul 6, 2006
Messages
6,857
Likes
87
Location
Rotterdam (actually Capelle aan den IJssel), the N
WCA
2006GALE01
YouTube
arnaudvg
#25
imagine knowing all of god's algorithms, that guy would be amazing at blind
Only if he could remember the alg quick enough and execute it quick enough.
For a computer/robot it would probably be faster to calculate and execute a non-optimal 21 move alg then to calculate and execute an optimal 19 move alg
 
Last edited:
Joined
Oct 31, 2008
Messages
269
Likes
14
#26
For a set of 1000 random positions, the two-phase algorithm on a recent PC,
given 100ms to work with, can calculate on average a solution of length 19.
So a lot depends on how fast your robot is.
 
Joined
Jul 25, 2006
Messages
880
Likes
0
Location
0, 0
#27
Does anyone know how to calculate the average for color neutral non-matching CLL (R', R, and R2 layers) versus CLL and color neutral non-matching EG (R', R, and R2 faces) versus EG?
 
Joined
Oct 1, 2010
Messages
361
Likes
0
#28
Does anyone know how to calculate the average for color neutral non-matching CLL (R', R, and R2 layers) versus CLL and color neutral non-matching EG (R', R, and R2 faces) versus EG?
Isn't the average of neutral non-matching CLL equal to (average of CLL) + 1.5 (1.5 from Aufing R and then U ) ?

What's the difference for the first layer ?
 
Joined
Jul 25, 2006
Messages
880
Likes
0
Location
0, 0
#32
Isn't the average of neutral non-matching CLL equal to (average of CLL) + 1.5 (1.5 from Aufing R and then U ) ?

What's the difference for the first layer ?
Yep. The only information I'm wanting is layer versus non-matching layer and face versus non-matching face.
 

Lucas Garron

Moderator
Staff member
Joined
Jul 6, 2007
Messages
3,555
Likes
88
Location
California
WCA
2006GARR01
YouTube
LucasGarron
#33
Great video! This should have its own thread (or maybe be in the 20-thread).

(and great for me to finally see Tom and John on video (I've known them for ~9 years from TopCoder))
Hmm, I randomly seeked the video and saw 43,000,000,000,000,000,000. :(

But the transcript looks better. Will try to watch when I'm done coding.

EDIT: Just watched it. Apart from that single part, looks like the best cubing TV report I've ever seen; thorough and accessible. It's too bad this would never spread as fast as fake news. :-/
 
Last edited:
Joined
Oct 8, 2006
Messages
914
Likes
15
Location
Malden, MA, USA
WCA
2006NORS01
YouTube
cuBerBruce
#35
God's number for a pure 5-cycle of edges?

God's number for a pure 5-cycle of corners?

It would be even more awesome to get the move-count distribution :p
All corner 5-cycles can be solved in 15 or less face turns.

The distribution in terms of symmetry/antisymmetry equivalence classes is given below. Note that the equivalence classes are not all the same size so you can't calculate the exact average from these numbers.

Cube Explorer was used to generate this distribution.

Cubes solved optimally: 1152

10f*: 13
11f*: 31
12f*: 229
13f*: 445
14f*: 414
15f*: 20

EDIT:
For edge 5-cycles, again up to 15 face turns may be required. The distribution (again reduced by symmetry/antisymmetry) is below.

Cubes solved optimally: 3272

6f*: 3
7f*: 5
8f*: 23
9f*: 57
10f*: 248
11f*: 579
12f*: 1212
13f*: 1011
14f*: 132
15f*: 2

Shortest maneuvers include:
L' R B L R' U' (6f*)
B' U' B F' L F (6f*)
L U' L' R B R' (6f*)

Longest optimal maneuvers include:
U F R U2 F' R B F' U' B' F2 U2 R' F' U' (15f*)
L2 D U' F2 R U2 F' U2 R2 U2 F' D2 L D U' (15f*)
 
Last edited:
Joined
Dec 11, 2009
Messages
294
Likes
0
#37
All corner 5-cycles can be solved in 15 or less face turns.

The distribution in terms of symmetry/antisymmetry equivalence classes is given below. Note that the equivalence classes are not all the same size so you can't calculate the exact average from these numbers.

Cube Explorer was used to generate this distribution.

Cubes solved optimally: 1152

10f*: 13
11f*: 31
12f*: 229
13f*: 445
14f*: 414
15f*: 20
For 5-corners located at (1) DFR and (4) in U-layer?
And then, not counting any equivalencies that can be achieved by AUF?
 
Joined
Oct 8, 2006
Messages
914
Likes
15
Location
Malden, MA, USA
WCA
2006NORS01
YouTube
cuBerBruce
#38
For 5-corners located at (1) DFR and (4) in U-layer?
And then, not counting any equivalencies that can be achieved by AUF?
The full distribution of 5-cycles for these 5 corners is given below. Since this is not a symmetry-reduced distribution, the exact average can be calculated. It is 13 + 67/486 or approximately 13.1379.

Cubes solved optimally: 1944

10f*: 24
11f*: 52
12f*: 372
13f*: 716
14f*: 744
15f*: 36

For the 2nd part, I assume you want these 5-cycles grouped in sets of 4 through conjugation by AUF moves, and only counting the shortest maneuver for each set of four configurations.

EDIT: Based upon the above assumptions, I've rerun the 1944 with the sets of 4 cases grouped together, and used a perl script to extract the best of each set of 4 configurations. The result I get is:

10f*: 16
11f*: 20
12f*: 186
13f*: 212
14f*: 52

Total cases: 486

By using conjugations without bothering to count the setup and undo setup moves, we get a "better than optimal" distance distribution. :)
 
Last edited:
Joined
Jun 17, 2006
Messages
654
Likes
1
#39
God's number for inserting edge/corner 5-cycle into a skeleton?? Hehe ... Move cancellations is the issue here!!

Per
 
Last edited:
Top