Number of Cube States With No Same-Colored Stickers Adjacent?

[Dapianokid]

Is this calculatable? If so, has it been calculated? If so, do tell, how many different positions are there of a cube in which there aren't any adjacent stickers? and what are the odds that a humanly scrambled cube could begin in a state with no adjacent stickers? My friend and I had a competition to see who could get a scrambled cube to a position without adjacency the fastest without at all solving it (He knew I could solve it into a multicolored cross position without having to actually solve it.) and it took us both a long time! I know that misorienting all 12 edges in a position like Superflip would get you there, but I want to know how many there are! I'm guessing very few.

 

[cubernya]

Is this calcuable? If so, has it been calculated? If so, do tell, how many different positions are there of a cube in which there aren't any adjacent stickers? and what are the odds that a humanly scrambled cube could begin in a state with no adjacent stickers? My friend and I had a competition to see who could get a scrambled cube to a position without adjacency the fastest without at all solving it (He knew I could solve it into a multicolored cross position without having to actually solve it.) and it took us both a long time! I know that misorienting all 12 edges in a position like Superflip would get you there, but I want to know how many there are! I'm guessing very few.

Calculable? Yes, but not in pure math (I think).
Calculated? No
Odds of a human scrambled with that: Impossible to calculate, since humans don't scramble consistently.

 

[cuBerBruce]

It appears you want to know the number of positions that have no adjacent (excluding diagonal adjacency) stickers being of the same color.

Well, each edge sticker is adjacent to 2 corner stickers and 1 center sticker. We can expect the probability of each such adjacent pair of stickers to have non-matching colors to be 5/6. Assuming these probabilities to be independent (a false assumption, but should be a "good enough" assumption to give a crude approximation of the correct probability), we get a probability of (5/6)^(24*3) which is approximately 1 in 502400. This would imply the actual number to be about 86 trillion positions.

Calculating the exact answer I would guess to be a very involved calculation, and probably take a very considerable amount of computer time.

 

[qqwref]

I agree with Bruce - this is not easy to calculate and I don't think it has been done. Given the raw size of the problem I can't think of any good way to do it either.

 

[Dapianokid]

To clear up terminology: No adjacent colored stickers. E.g. Red next to red is a no no. There are also no states, I believe, without adjacencies including diagonals, because of the properties of the orientations of corners. Diagonals are okay. Bruce is onto something.

 

[MaikeruKonare]

I know of like 6, checkerboard, flower, flower cross, and fancy checkerboard. Examples
M2 E2 S2
M E' M' E M2 E2 S2
F B2 R' D2 B R U D' R L' D' F' R2 D F2 B'
Then there are variations of the bottom two.

 

[Patrick M]

2x2 can have no adjacent stickers of same color.

 

[MaikeruKonare]

2x2 can have no adjacent stickers of same color.

Don't you hate those cases? It take longer to build a face. Two inserts instead of a slide + insert.

 

[qqwref]

There are also no states, I believe, without adjacencies including diagonals, because of the properties of the orientations of corners. Diagonals are okay. Bruce is onto something.

S M2 E'

 

[MaikeruKonare]

S M2 E'

Pro. So easy too

 

[Patrick M]

Don't you hate those cases? It take longer to build a face. Two inserts instead of a slide + insert.

Usually yes even if i use ortega, but one time i got an easy first layer + ll skip so maybe i reversed solved lol

 

[irontwig]

Katsuyuki Konishi once played around with these kind of scrambles:

http://www.planet-puzzle.com/cubekyukan2x2.html
http://www.planet-puzzle.com/cubekyukan3x3.html
http://www.planet-puzzle.com/cubekyukan4x4.html
http://www.planet-puzzle.com/cubekyukan4x4b.html
http://www.planet-puzzle.com/5x5x5/5x5x5_method/5x5x5_algorithm.html
http://www.planet-puzzle.com/syasinkan.html

 

[Stefan]

Don't you hate those cases? It take longer to build a face. Two inserts instead of a slide + insert.

Scramble: F' R2 U
Solve: U' R2' F

S M2 E'

FLU and LF are the same color and are kinda diagonally adjacent. Can you get rid of that, too?

 

[Kirjava]

Scramble: F' R2 U
Solve: U' R2' F

isn't this now an invalid scramble though?

 

[qqwref]

FLU and LF are the same color and are kinda diagonally adjacent. Can you get rid of that, too?

As far as I can tell, no pattern that looks like this (corners solved, opposite edges on a face the same color, 4 colors per face) can also satisfy that non-adjacency criterion.

 

[Lchu613]

Hmm. Interesting math problem. Tell me when somebody decides to have a go at it.

Also are you referring too possible cube states by scramble from solved state, or do you include unsolvable states?

 

[Dapianokid]

Lchu, I don't intend to include unsolvable states. That's mostly because it adds undesirable complexity to the problem.
My conjecture about non-diagonals was disproved rather quickly. But I do love the math behind cubes and I would love us to all seriously consider it. the 2x2x2 may be a better cube to try this with, as it has fewer states and no centers to worry about.

 

[qqwref]

FLU and LF are the same color and are kinda diagonally adjacent. Can you get rid of that, too?

As far as I can tell, no pattern that looks like this (corners solved, opposite edges on a face the same color, 4 colors per face) can also satisfy that non-adjacency criterion.

I think I've done it. I kept the "corners solved" criterion but relaxed the others. In the following pattern, no two stickers that are adjacent, directly or diagonally, even across the boundaries of the cube's faces, have the same color.
B' L' D' L' F' R' D' B' U' F' D' L R2 D U' B' U' (17f*)
y2 x' D2 B' D2 L S R' B2 L' E2 R2 E L' S U R2 (15s*)

 

[Dapianokid]

The number of Cube states in a 2x2x2 with no same colored stickers adjacent is most likely comparable to the number of legal states of the cube.

 

[cuBerBruce]

For 2x2x2, my program got a count of 146 2x2x2 positions for which no same-color sticker are adjacent, not even diagonally whether or not across a face boundary. An example:

F U' R2 F2 U2 R' F'

I've previously reported counting 17004 2x2x2 positions with four different colored stickers on each face in [thread=25272]this thread[/thread].