# How many solved positions for 3x3x3?



## yboy403 (Mar 24, 2010)

I know the standard numbers are 43,252,003,274,489,856,000 possible configurations and 1 solved position. However, I was thinking about it and realized that if a cube was solved except for one move, that could basically be called a solved position, because most speedcubers could solve it in less than 0.1 seconds. Same goes for positions requiring two turns to solve, those aren't much harder.
Agree? Disagree? Is there any way to include those in the tally of "solved" positions? Please post.
(My first thread)


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## koreancuber (Mar 24, 2010)

Would it be infinite? And one turn is not the solved position, it's one move *from* the solved position.


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## dannyz0r (Mar 24, 2010)

1move +2


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## TioMario (Mar 24, 2010)

WCA says solved is solved... period .
One turn from solved is +2


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## iSpinz (Mar 24, 2010)

yboy403 said:


> Because most speedcubers could solve it in less than 0.1 seconds.


Really?


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## ianini (Mar 24, 2010)

I count 19. That's if R and R' as +'2 count as separate.


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## mr. giggums (Mar 24, 2010)

Well if you consider a +2 a solved state then count the ways:
U U2 U' R R2 R' L L2 L' F F2 F' B B2 B' D D2 D'
Thats 18 plus the solve state so 19 in 43,252,003,274,489,856,000 would be considered solved.


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## ianini (Mar 24, 2010)

^^ That's what I thought.


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## DaijoCube (Mar 24, 2010)

I love mathematics <3


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## Cyrus C. (Mar 24, 2010)

That was more like counting, not (tricky) mathematics.


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## cmhardw (Mar 24, 2010)

Exactly 2 turns away from solved I count:

18*15 - 3*3*3 = 243

There are 18 possible first turns. After which there are 15 possible turns of a *different* layer (5 layers to choose from, and each layer has 3 ways to spin). However, this doubles counts all possible turns of two parallel layers. Considering any two parallel layers there are 3 ways to spin the first layer and 3 ways to spin the parallel layer. There are 3 sets of parallel layers, so 3*3*3 = 27 positions have been double counted.

So I guess you could say that counting the solved state, 1 move away from solved, and exactly 2 moves away from solved there are:

1 + 18 + 243 = 262 "solved" positions based on the first part of your question.

Chris


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## miniGOINGS (Mar 24, 2010)

iSpinz said:


> yboy403 said:
> 
> 
> > Because most speedcubers could solve it in less than 0.1 seconds.
> ...



My one move scramble for 2x2 was exactly 0.10.


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## Cyrus C. (Mar 24, 2010)

What about 2 turns away? Most cubers can do 2 turns in less than .5 seconds.


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## yboy403 (Mar 24, 2010)

iSpinz said:


> yboy403 said:
> 
> 
> > Because most speedcubers could solve it in less than 0.1 seconds.
> ...


hell, yeah. 0.1 seconds for a R move? Easy. If you can do a sune in 0.7, then work the rest out yourself.


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## yboy403 (Mar 24, 2010)

cmhardw said:


> Exactly 2 turns away from solved I count:
> 
> 18*15 - 3*3*3 = 243
> 
> ...


doesn't have the same ring to it, does it? 
"43 quadrillion possible positions, and only 262 solved positions"


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## ben1996123 (Mar 24, 2010)

I don't count 1 move away from solved as being solved.

My answer would have to be 2,048, due to centre rotations.


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## yboy403 (Mar 24, 2010)

koreancuber said:


> Would it be infinite? And one turn is not the solved position, it's one move *from* the solved position.





dannyz0r said:


> 1move +2





TioMario said:


> WCA says solved is solved... period .
> One turn from solved is +2





ben1996123 said:


> I don't count 1 move away from solved as being solved.
> 
> My answer would have to be 2,048, due to centre rotations.



Interesting. anyway, this has nothing to do with _speed_cubing or the WCA, just cubing in general. What I'm saying is that when Rubik's prints on their packages, or Dan Harris writes on the back of his book (good book BTW), that the ratio of total positions to solved positions is 43,252,003,274,489,856,000:1, they're leaving out the fact that there are an unspecified number (ben1996123 says 2048, Chris says around 300) of positions that a person with no knowledge of Rubik's Cube could solve easily, and which most people (WCA aside) would not consider a scrambled state. And even the WCA acknowledges that one turn from solved is _something_, because they only give a penalty for it, not a DNF. All I want to know is, including up to three turns from solved, which is the maximum my 5-year-old sister has solved with no help, how many unstated "solved" positions are there besides the standard one with all sides the same colour. If you do come up with a number, I'd like to see how you got it, purely for educational purposes (my own)
Yerachmiel


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## cuBerBruce (Mar 24, 2010)

ben1996123 said:


> I don't count 1 move away from solved as being solved.
> 
> My answer would have to be 2,048, due to centre rotations.



Good answer... except then the total number of configurations then has to be counted as 88580102706155225088000. This is really the correct number of (cubic) configurations of the pieces (assuming we don't care how the cube is oriented in space).


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## DavidWoner (Mar 24, 2010)

1 move from solved is not solved. Period.



yboy403 said:


> iSpinz said:
> 
> 
> > yboy403 said:
> ...



You have to start and stop the timer.


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## yboy403 (Mar 24, 2010)

DavidWoner said:


> 1 move from solved is not solved. Period.
> 
> 
> 
> ...


see his time for these cubes. Probably if you had to start and stop a timer it would be something similar. anyway, read my post two up from yours (above CuBerBruce). This really has nothing to do with speed, just difficulty. I just gave a rough estimate of the time needed to solve it (no timers involved) as an idea of how easy it is, and how it's more of a solved position than a scrambled one.
Yerachmiel


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## miniGOINGS (Mar 24, 2010)

The problem isn't doing the R, it's the timer.


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## yboy403 (Mar 24, 2010)

miniGOINGS said:


> The problem isn't doing the R, it's the timer.



Yeah, DavidWoner said that. That's the point: for almost all intents and purposes, except for speedcubing, one R from solved _is_ solved.


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## miniGOINGS (Mar 24, 2010)

yboy403 said:


> Yeah, DavidWoner said that. That's the point: for almost all intents and purposes, except for speedcubing, one R from solved _is_ solved.



I just saw that. I would still be careful how you word it though.


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## yboy403 (Mar 24, 2010)

miniGOINGS said:


> yboy403 said:
> 
> 
> > Yeah, DavidWoner said that. That's the point: for almost all intents and purposes, except for speedcubing, one R from solved _is_ solved.
> ...



?


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## miniGOINGS (Mar 24, 2010)

yboy403 said:


> miniGOINGS said:
> 
> 
> > yboy403 said:
> ...



That's what I'm talking about.


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## cuBerBruce (Mar 24, 2010)

yboy403 said:


> koreancuber said:
> 
> 
> > Would it be infinite? And one turn is not the solved position, it's one move *from* the solved position.
> ...



If you allow a third turn from the solved position, you can reach 3240 additional positions. That added to the 262 positions of reachable in two moves or less makes a total of 3502 positions. A computer algorithm called a breadth-first search can be used to calculate these numbers. Such numbers have been computed out to 13 moves (face turns). See: God's Algorithm out to 13f*.

Chris worked this out by hand for two moves, and it is not too difficult to extend that type of argument to three moves to come up with the answer 3502. It perhaps just gets a little trickier to figure out all the duplicates. Going out further than 3 moves you can start reaching duplicate positions in "unexpected" ways, making it even trickier to avoid overcounting.


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## yboy403 (Mar 24, 2010)

miniGOINGS said:


> yboy403 said:
> 
> 
> > miniGOINGS said:
> ...



With no disrespect intended, do you think you might manage to write a complete sentence explaining what you mean? Do you mean that I shouldn't say it's _actually_ solved?


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## CuBeOrDiE (Mar 24, 2010)

well, if its solved or not depends on the stickers. if there are none, then there are about 43 quintillion solved positions.


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## miniGOINGS (Mar 24, 2010)

yboy403 said:


> With no disrespect intended, do you think you might manage to write a complete sentence explaining what you mean?



I was just thinking that if that point wasn't written out like it is, it might be misleading. I have no problem with "for almost all intents and purposes, except for speedcubing, one R from solved is solved". But if it was worded differently, like "one R away from solved is still solved" it's a very different statement. At least in my opinion.


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## yboy403 (Mar 24, 2010)

miniGOINGS said:


> yboy403 said:
> 
> 
> > With no disrespect intended, do you think you might manage to write a complete sentence explaining what you mean?
> ...


Now I get it. Thanks. I'll try and be clearer in future.


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## miniGOINGS (Mar 24, 2010)

yboy403 said:


> Now I get it. Thanks. I'll try and be clearer in future.



I don't think you have to be any clearer, just try not to get sloppy.


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## yboy403 (Mar 24, 2010)

CuBeOrDiE said:


> well, if its solved or not depends on the stickers. if there are none, then there are about 43 quintillion solved positions.


Yes, but it's also approximately the same difficulty as a 1x1x1. Where's the fun in that (besides reviewing an unstickered cube on YouTube)?


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## yboy403 (Mar 24, 2010)

miniGOINGS said:


> yboy403 said:
> 
> 
> > Now I get it. Thanks. I'll try and be clearer in future.
> ...



k. I know that's one of my problems: generalizations, overlooking obvious points, etc.
Yerachmiel


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## inuyasha51089 (Mar 24, 2010)

anyone ever try to olve a cubeto the  checker board patteren lol it adds like 5-10 seconds to my solves but its funn


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## andyt1992 (Mar 24, 2010)

centers can rotate.


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## mr. giggums (Mar 24, 2010)

CuBeOrDiE said:


> well, if its solved or not depends on the stickers. if there are none, then there are about 43 quintillion solved positions.



actually there is one position because you would be able to tell if it were a R the away they just look the same just like on a 4x4x4 and up the centers can be arranged any way and still be solved as long as the're on the right sides.


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## yboy403 (Mar 24, 2010)

andyt1992 said:


> centers can rotate.



How very true. I once solved my 3x3 four times, taking note of the orientation of the Rubik's logo on the white face. It faced a different way after each solve. Is that what you meant?


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## cuBerBruce (Mar 25, 2010)

It's also what ben1996123 was referring to (if you didn't realize that) many posts back. On a 3x3x3, this multiplies the number of positions by (4^6)/2 = 2048 times. There is a divide by 2 above because of a parity constraint.

A cube stickered so that the rotational orientation of the centers matter (and positions of the non-central face pieces on a 4x4x4 or larger cube) is generally called a supercube.


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## kunz (Mar 25, 2010)

would you count the centers flipping (like in picto cubes) as a different position?


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## yboy403 (Mar 25, 2010)

kunz said:


> would you count the centers flipping (like in picto cubes) as a different position?



I guess that's what I mean. When the flip of the centres affects whether or not the cube is 'solved', the number of positions basically quadruples. However, in a solid-coloured or -patterned cube, where the center rotation makes little practical difference, those extra positions are only theoretical. So, the answer is that it depends on the cube.
Yerachmiel


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## cuBerBruce (Mar 25, 2010)

yboy403 said:


> kunz said:
> 
> 
> > would you count the centers flipping (like in picto cubes) as a different position?
> ...



Well, it quadruples if only one center's orientation matters. If all 6 matter, the number of position increases 2048 times, as I've said more than once I think.


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## Jun Kim (Mar 25, 2010)

you have to add 2 seconds


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## yboy403 (Mar 25, 2010)

cuBerBruce said:


> yboy403 said:
> 
> 
> > kunz said:
> ...



Very true. I was thinking of my Rubik's.com cube, which only has one center design, but on a pictocube you're right, all the centres would matter,
Yerachmiel.


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## yboy403 (Mar 25, 2010)

Jun Kim said:


> you have to add 2 seconds



Why? The timer? If that's what you're thinking of, read the rest of the thread first.
Yerachmiel


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## shelley (Mar 25, 2010)

There are 43 quintillion possible positions. That doesn't mean there are 43 quintillion positions in which the cube is completely scrambled. A cube that is one or two moves away from solved is NOT SOLVED. It counts in one of the 43 quintillion positions that is not solved. Why is there this much discussion?

If you wanted to know how many possible positions are one or two moves away from solved, you should ask that, don't try to redefine the solved state.

And to the person who said there are infinite positions... you're an idiot.


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## yboy403 (Mar 26, 2010)

shelley said:


> There are 43 quintillion possible positions. That doesn't mean there are 43 quintillion positions in which the cube is completely scrambled. A cube that is one or two moves away from solved is NOT SOLVED. It counts in one of the 43 quintillion positions that is not solved. Why is there this much discussion?
> 
> If you wanted to know how many possible positions are one or two moves away from solved, you should ask that, don't try to redefine the solved state.
> 
> And to the person who said there are infinite positions... you're an idiot.



Wow, I guess you're just smarter than all of us (insert heavy sarcasm here).
If you'd actually read the thread, you'd realize that although the thread title _is_ misleading, because I wrote it in a hurry, there are lines like:
Is there any way to include those in the tally of "solved" positions? (Note the quotes around "solved")

Also (the killer):


yboy403 said:


> Interesting. anyway, this has nothing to do with speedcubing or the WCA, just cubing in general. What I'm saying is that when Rubik's prints on their packages, or Dan Harris writes on the back of his book (good book BTW), that the ratio of total positions to solved positions is 43,252,003,274,489,856,000:1, they're leaving out the fact that there are an unspecified number (ben1996123 says 2048, Chris says around 300) of positions that a person with no knowledge of Rubik's Cube could solve easily, and which most people (WCA aside) would not consider a scrambled state. And even the WCA acknowledges that one turn from solved is something, because they only give a penalty for it, not a DNF. All I want to know is, including up to three turns from solved, which is the maximum my 5-year-old sister has solved with no help, how many unstated "solved" positions are there besides the standard one with all sides the same colour. If you do come up with a number, I'd like to see how you got it, purely for educational purposes (my own)
> Yerachmiel


It's clearly stated there that I'm looking for 


yboy403 said:


> [P]ositions that a person with no knowledge of Rubik's Cube could solve easily, and which most people (WCA aside) would not consider a scrambled state.


That's basically the definition of what this thread is about. 
*For everybody reading this: This thread is not talking about solved positions. Read the rest of this post before you reply.* If you'd bothered to read it, you'd know that. This isn't about speed, it's about difficulty, in terms of which one or two turns from solved is barely more difficult than being solved, besided the physical effort of turning. That's also why this thread is in the theory section and not the speedcubing section. 
anyway, back to the original topic. While I agree that there are clearly not an infinite number of positions, posting on a thread you haven't read completely is a surer sign of idiocy than a statement that happens to contain a mistake (which we all make). The reason there's so much discussion is because it's an interesting (at least to me) discussion about the difficulty and mathematics of Rubik's Cube.
Yerachmiel


Spoiler



Shelley: As with all my posts, if you find any of this offensive to you, tell me and I will edit it out


Sorry this post is so long, it's just that there's been a lot of confusion over it and I just wanted to clear it up.


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## shelley (Mar 26, 2010)

I did read the thread. Your question was answered by Chris on the first page. The rest of it seemed like an unnecessarily long argument over your usage of the term "solved state" when that could have been easily avoided by saying what you meant in the first place.


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## yboy403 (Mar 26, 2010)

shelley said:


> I did read the thread. Your question was answered by Chris on the first page. The rest of it seemed like an unnecessarily long argument over your usage of the term "solved state" when that could have been easily avoided by saying what you meant in the first place.


That's true. But then I asked a new question, "What about 3 turns?" That one seems to have been lost in the confusion over the my bad wording.
Yerachmiel


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## cmhardw (Mar 26, 2010)

yboy403 said:


> That's true. But then I asked a new question, "What about 3 turns?" That one seems to have been lost in the confusion over the my bad wording.
> Yerachmiel




Unless I misinterpreted your question, Bruce already answered that.



cuBerBruce said:


> If you allow a third turn from the solved position, you can reach 3240 additional positions. That added to the 262 positions of reachable in two moves or less makes a total of 3502 positions. A computer algorithm called a breadth-first search can be used to calculate these numbers. Such numbers have been computed out to 13 moves (face turns). See: God's Algorithm out to 13f*.
> 
> Chris worked this out by hand for two moves, and it is not too difficult to extend that type of argument to three moves to come up with the answer 3502. It perhaps just gets a little trickier to figure out all the duplicates. Going out further than 3 moves you can start reaching duplicate positions in "unexpected" ways, making it even trickier to avoid overcounting.




Chris


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## yboy403 (Mar 26, 2010)

Must have missed that one. 
I think this thread is getting too confusing. I have my answer already. Anybody else want to kill it?
Yerachmiel


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## cuBerBruce (Mar 26, 2010)

cmhardw said:


> yboy403 said:
> 
> 
> > That's true. But then I asked a new question, "What about 3 turns?" That one seems to have been lost in the confusion over the my bad wording.
> ...



I gave the number. I didn't go through how you derive it manually.

Basically you go through a procedure like Chris did for 2 moves. To avoid overcounting I will simply avoid having a U-layer turn following a D-layer turn, avoid an L-layer turn following an R-layer turn, and avoid an F-layer turn following a B-layer turn. The table below then gives the possibilities when the first move is a U- or D-layer turn. Each letter represents three possible turns of the indicated layer. 


```
1st  2nd  3rd     count (for 3rd move)

U     D   LRFB    4*3
U     L   UDRFB   5*3
U     R   UDFB    4*3
U     F   UDLRB   5*3
U     B   UDLR    4*3
D     L   UDRFB   5*3
D     R   UDFB    4*3
D     F   UDLRB   5*3
D     B   UDLR    4*3
```
The first and 2 moves have 3 choices for each case (3 ways to rotate the layer) so the total count is 3*3*(4*3+5*3+4*3+5*3+4*3+5*3+4*3+5*3+4*3) = 1080.

We can do the same thing for the first turn being on L/R, and for the first turn being on F/B. This triples the number to 1080*3=3240.

(Be careful applying this procedure to more than 3 moves, because you will start reaching duplicate positions in non-obvious ways.)


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## yboy403 (Mar 26, 2010)

Thanks. I wasn't planning to do more than 3 moves anyway, because most non-cubers would probably make at least one wrong move if they didn't inspect it first for a long time. _Now_ this thread's purpose has been served. You guys can continue it if you want, but I'm gone.
Thanks everybody,
Yerachmiel


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## FatBoyXPC (Mar 26, 2010)

shelley said:


> There are 43 quintillion possible positions. That doesn't mean there are 43 quintillion positions in which the cube is completely scrambled. A cube that is one or two moves away from solved is NOT SOLVED. It counts in one of the 43 quintillion positions that is not solved. Why is there this much discussion?
> 
> If you wanted to know how many possible positions are one or two moves away from solved, you should ask that, don't try to redefine the solved state.
> 
> And to the person who said there are infinite positions... you're an idiot.



I was going to say nearly this exact thing, including the 43 quintillion possibilities. No offense to anybody, but I don't think a person of average intelligence could solve a cube that is 3 turns away. I realize this seems rather easy, but I've done this exact thing with many many people who said something to the effect of "Wow I couldn't solve it unless it was like less than 5 moves." I'll usually give them 4, when they can't do it I give them 3, they still usually can't do it. Don't get me wrong, I've seen people do 3 and even 4 move scrambles, but the overall average is they can't do it with 3 turns.


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## yboy403 (Mar 26, 2010)

fatboyxpc said:


> I'll usually give them 4, when they can't do it I give them 3, they still usually can't do it. Don't get me wrong, I've seen people do 3 and even 4 move scrambles, but the overall average is they can't do it with 3 turns.



I don't know about the people you hang around with, but I've seen my 5-year-old sister do three moves several times in a row with no help-three random moves, too.


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## iasimp1997 (Mar 26, 2010)

One.


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## yboy403 (Mar 26, 2010)

iasimp1997 said:


> One.



??


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## TeddyKGB (Mar 27, 2010)

This may be a few days late, but I thought I would give my opinion, I dont think one move away from solved would be considered solve, it wouldn't nessisarily be scrambled but its not solved! maybe "slightly askew" so the real question is "how many slightly askew positions are there?"


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## megaminxwin (Mar 27, 2010)

mr. giggums said:


> Well if you consider a +2 a solved state then count the ways:
> U U2 U' R R2 R' L L2 L' F F2 F' B B2 B' D D2 D'
> Thats 18 plus the solve state so 19 in 43,252,003,274,489,856,000 would be considered solved.



Don't forget the x24


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## yboy403 (Mar 28, 2010)

TeddyKGB said:


> This may be a few days late, but I thought I would give my opinion, I dont think one move away from solved would be considered solve, it wouldn't nessisarily be scrambled but its not solved! maybe "slightly askew" so the real question is "how many slightly askew positions are there?"



Right, which is why I've said a number of times that all the confusion is over my bad wording of the thread title.
Yerachmiel


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