How many solved positions for 3x3x3?

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

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

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 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

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.
 
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.

?

That's what I'm talking about.

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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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.
 
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.
Now I get it. Thanks. I'll try and be clearer in future.
 
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.
 
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.
 
would you count the centers flipping (like in picto cubes) as a different position?
 
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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