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



## Dapianokid (Aug 26, 2013)

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.


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## cubernya (Aug 26, 2013)

Dapianokid said:


> 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.


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## cuBerBruce (Aug 26, 2013)

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.


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## qqwref (Aug 27, 2013)

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.


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## Dapianokid (Aug 27, 2013)

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.


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## MaikeruKonare (Aug 27, 2013)

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.


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## Patrick M (Aug 27, 2013)

2x2 can have no adjacent stickers of same color.


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## MaikeruKonare (Aug 27, 2013)

Patrick M said:


> 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.


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## qqwref (Aug 27, 2013)

Dapianokid said:


> 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'


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## MaikeruKonare (Aug 27, 2013)

qqwref said:


> S M2 E'



Pro. So easy too


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## Patrick M (Aug 27, 2013)

MaikeruKonare said:


> 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


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## irontwig (Aug 27, 2013)

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


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## Stefan (Aug 27, 2013)

MaikeruKonare said:


> 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



qqwref said:


> S M2 E'



*F*LU and *L*F are the same color and are kinda diagonally adjacent. Can you get rid of that, too?


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## Kirjava (Aug 27, 2013)

Stefan said:


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



isn't this now an invalid scramble though?


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## qqwref (Aug 27, 2013)

Stefan said:


> *F*LU and *L*F 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.


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## Lchu613 (Aug 27, 2013)

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?


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## Dapianokid (Aug 27, 2013)

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.


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## qqwref (Aug 27, 2013)

qqwref said:


> Stefan said:
> 
> 
> > *F*LU and *L*F are the same color and are kinda diagonally adjacent. Can you get rid of that, too?
> ...


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*)


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## Dapianokid (Aug 28, 2013)

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.


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## cuBerBruce (Aug 28, 2013)

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].


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## Dapianokid (Aug 28, 2013)

And did this test every case? and counting across face boundaries significantly lowers the count of states and is harder for a human to check. The reason I asked originally is because my friends generally go for states with no adjacencies. I have to tell them that it's about as hard as actually solving the cube. My mom often randomly gets scrambles that satisfy the rule about no same colored sticker adjacencies, so she is beating incredible odds without trying.


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## cmhardw (Aug 28, 2013)

Dapianokid said:


> My mom often randomly gets scrambles that satisfy the rule about no same colored sticker adjacencies, *so she is beating incredible odds without trying.*



I really doubt this to be true. I don't think your mom is scrambling completely randomly without examining the cube at all. If she does that, and still gets the no-adjacencies scrambles, then this is beating incredible odds. I'll bet your mom is actively trying to break up blocks of the same color. This will drastically change the odds.

As a comparison, imagine you scramble a cube with your eyes closed, then open them up and see if you have a solved cube. Chances are 1 in 43252003274489856000 that you do have a solved cube.

Now, solve the first two layers of your cube. Do the last block/pair with your eyes closed, then open them. Chances are 1/62208 that you have a solved cube. Actively giving the cube a little bit of order will drastically improve the probability of having a solved cube by chance.

In a similar way your mom drastically improves her chances of a no-adjacencies case by getting 2 or 3 sides to have no adjacent stickers of the same color.


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## Dapianokid (Aug 28, 2013)

She also tends to do very few moves on the cubes. So, she isn't really decreasing the order of the cube by an insane amount. 
She told me it was all by chance. I think she's playing a joke. I wouldn't say she was incredibly lucky to get attention  but I seriously don't know how she does it. I have a hard time even scrambling the cube to such positions and she does it differently every time in 10 moves or so! Is there a way of quickly identifying how to create no adjacencies?


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## Kirjava (Aug 28, 2013)

cuBerBruce said:


> F U' R2 F2 U2 R' F'



omg that sucks


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## Dapianokid (Aug 29, 2013)

I'm guessing the number of positions that satisfy this rule are spread evenly accross the cube groups.


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## Stefan (Aug 29, 2013)

Dapianokid said:


> I'm guessing the number of positions that satisfy this rule are spread evenly accross the cube groups.



What does that mean?


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## Dapianokid (Aug 29, 2013)

That the distribution of these states is regular accross all size cubes and classified states. They appear as often with 2x2x2's as they do 3x3x3's as they do 4x4x4's.


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## cmhardw (Aug 29, 2013)

Dapianokid said:


> That the distribution of these states is regular accross all size cubes and classified states. They appear as often with 2x2x2's as they do 3x3x3's as they do 4x4x4's.



Are you saying that the proportion of states on a 2x2x2 that don't have adjacent stickers of the same color to the total number of 2x2x2 states is the same as the proportion of states on a 3x3x3 that don't have adjacent stickers of the same color to the total number of 3x3x3 states (and the same proportion on a 4x4x4, 5x5x5, etc.)?

This does not seem at all intuitive to me that such a thing would be true. Do you have a proof or at least line of reasoning that would lead you to this conclusion?


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## Dapianokid (Aug 29, 2013)

cmhardw said:


> Are you saying that the proportion of states on a 2x2x2 that don't have adjacent stickers of the same color to the total number of 2x2x2 states is the same as the proportion of states on a 3x3x3 that don't have adjacent stickers of the same color to the total number of 3x3x3 states (and the same proportion on a 4x4x4, 5x5x5, etc.)?
> 
> This does not seem at all intuitive to me that such a thing would be true. Do you have a proof or at least line of reasoning that would lead you to this conclusion?



It's obvious I need to word my posts better in this forum. 

I think it's proportional, but not in a linear sense. I think there is an eponential curve to the number of states that is constant. I think that there are more positions which satisfy this condition (and occur more often, as in they are easier to obtain) for a 4x4x4 than there are for a 3x3x3. I think the same for a 5x5x5 compared to a 4x4x4. I have no idea how to go about doing this math, though, but the interest is killing me.


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## cmhardw (Aug 29, 2013)

Dapianokid said:


> It's obvious I need to word my posts better in this forum.
> 
> I think it's proportional, but not in a linear sense.



If two things vary proportionately, then by definition they vary linearly. I don't understand what you mean when you say that things are proportional, but not linearly. Can you explain your thoughts further?



Dapianokid said:


> I think there is an eponential curve to the number of states that is constant. I think that there are more positions which satisfy this condition (and occur more often, as in they are easier to obtain) for a 4x4x4 than there are for a 3x3x3. I think the same for a 5x5x5 compared to a 4x4x4. I have no idea how to go about doing this math, though, but the interest is killing me.



I think what you're saying can be phrased as:

The probability of getting a no-adjacent stickers of the same color case on a 2x2x2 from a random scramble is the same as the probability of getting a no-adjacent stickers of the same color case on a 3x3x3 from a random scramble (and the same on a 4x4x4, 5x5x5, etc.).

Does that sound right?


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## Stefan (Aug 29, 2013)

Dapianokid said:


> I think there is an eponential curve to the number of states that is constant.



An eponential curve that is constant?


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## Dapianokid (Sep 1, 2013)

An exponential curve that remains consistently exponential.
I've decided to simply browse the forums until I have more cubing experience and more understanding of this place


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## Stefan (Sep 2, 2013)

Dapianokid said:


> An exponential curve that remains consistently exponential.



What is "consistently exponential"? And can you show an example of an exponential curve that isn't "consistently exponential"?


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## TMOY (Sep 2, 2013)

I don't know what he has in mind, but I can propose the folliowing mathematical definition of "consistently exponential":
A function \( f \) is consistently exponential if it is of the form \( f(x)=\lambda e^{\mu xg(x)} \), where \( \lambda \) and \( \mu \) are constants and \( g \) is a function whose limit in \( +\infty \) is 1.

For example, with this definition \( x\mapsto e^{x+\sqrt x} \) is consistently exponential but not exponential.


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