You can see 3 sides of a Rubik’s cube. The three sides all adjoin the same corner vertex — that is, as if you’re looking at the cube from above and to the left, and each visible side abuts the other two visible sides. The Cube is unsolved.
How many combinations could there be on the back three sides (assuming you know the color of each center square and nobody’s taken it apart or moved the stickers)?
I belive the answer is 24
There are 6 center pieces, but they never move relative to eachother, so we don’t worry about those. There are 8 corner pieces, but you can see 7 of them, so the one you can’t see can be rotated in 3 ways. There are 12 edge pieces, but you can see 9 of them, so the remaining 3 can be in 3 places and rotated in 2 ways.
There are 6 different ways the edges can be placed (ABC, ACB, BAC, BCA, CAB, CBA) and 8 ways for each placement that those can be rotated (for ABC: ABC, ABc, AbC, Abc, aBC, aBc, abC, abc). Thus, for each corner rotation, there are 48 possible edge configurations. Since there are 3 corner rotations, there should be 144 possible configurations. Since its not solve, then we subtract one.
My guess: 143
If you’ve got a standard Rubik’s cube you do indeed know what the colors in the centers of the other sides must be. In additiion, you know what the colors of the faces adjoining the cube faces you can see must be. That leaves only four “cubes” , representing three each having two faces (at the centers of the three unseen edges) that you can’t see, along with the one three-faced vertex cube you can’t see.
The fact that it’s a rubik cube, though, imits the number of possibilities. You can’t simply rotate that vertex cube to each of the three rotational positions and leave nothing else unchanged – only the one position is allowed. And you can’t, through any series of moves, only interchange two faces – you have to change at least three faces at a time. That also meansd you can’t “flip” one of those two-faced cubes to the other orientation.
I’m not sure how many “scramblings” that don’t change the observed faces are physically posible. I suspect it may only be one, although you might be able to interchange the three edge cubelets sequentially, which would give you a total of three. I can’t see how you could get as many as 24.
This is more complex without a cube in front of me. It occurs to me that each face has two esdge faces that I can only see one side of each. In principle, they could be interchanged and I’d never know (the other cubes on each of the faces I can see don’t have any freedom – they have to stay where they are). There are three faces , each of which have two cubes that can be interchanged.
BUT – you can’t interchange just two faces on a Rubik’s cube by manipulating it, so that doesn’t add 2 X 2 X 2 possibilities. You can interchanged three or four or more cubes at a time. Again, I’m not sure how many possibilities that adds, since I’m not sure how many of those interchanges are physically possible. At most there are four such changes that leave the cube looking as it did before – three involving four cubes at a time and one that involves all six edge cubes with one face showing. At the least, there’s only the one possibility.
So I still doubt that 24 is the number, although it’s possible. But I don’t see any way it can be as high as 143. I suspect the answer is one.
As others have done, this is easier by analyzing the individual cubelets. I’m assuming throughout that this is a standard Rubik’s cube which has gotten into whatever state it is by the usual twisting of faces, i.e. it hasn’t been taken apart and put back together in such a way that the cube is not longer solvable.
There is only one possibility for each of the center cubes; it is not possible to move these.
I can see seven of the eight corner cubes, and can deduce they are all correctly placed. Since it is impossible to twist a single corner cube by standard Rubiks manipulation without eventually twisting another cube in the exact opposite direction, we must conclude the eighth, unseen corner cube is placed properly (If you have a cube handy, try the following manipulation on a solved cube: F-, B, B, F, T, F-, B, B, F, where “F” indicates “rotate the front face clockwise”, “F-” = “rotate the front face counter-clockwise”. When completed, two of the corner cubelets on the top of the cube will have rotated; one 1/3 clockwise, one 1/3 counterclockwise).
The edges, then, are the only possible cubelets on the unseen sides that can be out of position. Six of these are partially unseen, three are completely unseen. The six partially unseen can be grouped in pairs with like color showing; these are each either correctly placed or swapped, and so there are 222=8 possibilities.
The three completely unseen edges, I think, can be placed in any of the three unseen edge positions, making for six possibilities. Finally, each of these edge piece can be in one of two orientations, but as with the corner pieces, you can’t flip one edge cubelet without flipping another in standard Rubik’s moves, limiting the total number of orientations to 4 (all 3 are correctly oriented, or only one of the three is correctly oriented). Thus, for the unseen edges, there are 6*4=24 possible orientations.
Multiply 648 = 192. It’s a back-of-the-envelope analysis, so I’d appreciate any clarifications/corrections. My guess is there is some other interaction between the position of the edge cubelets and their orientation which could limit this number further, but I don’t know what that could be.
CJJ and others – my argument is that a lot of those possibilities simply aren’t realizeable on a Rubik’s cube by standard manipulations. As I’ve said, I don’t think that any possibility of “switching” must involve three “cubelets” at minimum. This is not only my experience, but also the result of looking at books on “solving the cube”. No manipulation only interchanges two cubelets. So you can’t simply rotate one corner cubelet*, and I don’t believe you can simply interchange two corner cubelets. A lot of your possible configurations simply aren’t physically realizable.
so I might accept 24, but nothing over 100. And I only know for certain that one possibility is physically possible.
*If you want to drive someone nuts, disassemble a Rubik’s cube and re=assemble it with only one corner cube rotated out of position. Then scramble it. No one can ever solve it without disassembling the cube and putting one piece right.
This is rather complicated; the answer depends on the colors you can see.
You can see three centers, so you know the other three (up to orientation; if this is a standard Rubik’s cube the center orientation is not relevant, though). You can see all three sides of one corner, two sides of three more (allowing you to identify those three fully), and one side of three other corners; the remaining corner is hidden.
If the three faces you can see appear to be solved (e.g., you see pure red, green, and blue faces) then you can identify these three corners (since you know three of the four corners with that color already); and then the identity and orientation of the final corner are known (the corners are completely solved).
But if not, you can’t generally deduce the identities of all the corners. Suppose, for example, that the cube appears solved except that the color you can see on the back corner piece of the red face is white. This corner could be the red-white-yellow or orange-white-yellow corner; so there are two possible states for the corners.
(CalMeacham, you are correct that you can’t swap a single pair of edges or a single pair of corners; however, you can swap one pair of edges together with a single pair of corners.)
The situation for the edges is even more complicated. You can see three edges completely, and you can see one of two sides of six more; the remaining three edges are hidden. The back two edges on any of the three faces you see could be swapped; in addition, if the cube does not appear solved, there may be other edges you can’t determine.
Again assuming you see three apparently solved faces, there are eight possible permutations for the edges you see. The three remaining edges can be cyclically permuted (the total edge-plus-corner-permutation parity is even) and any two may be flipped; this gives another factor of 12. With the factor of 8 for the edges you see, this gives 96 possibilities. (Whew! I came in just under Cal’s threshold!)
So in the apparently-solved case, I count 96 possible states for the cube. In the general case you can’t deduce as much, so there are more possibilities.
That agrees with my statement about swapping three or more “cubelets”. But if you have a solved cube (which is what I’ve been assuming, although I now see that the OP doesn’t say that, so my numbers will be off if that’s not true) then you can’t swap the corners without it being obvious.
Hmm – I’ve been assumiong a solved cube (on the three visible sides). If the cube is random then there are a lot more possibilities.
You can deduce most of them. With the back corner, I agree: logically, on its own, there’s no knowing which of a few cubes it might be. However, the possibilities may be narrowed down by the three corners where two colors are visible, and the fourth corner where all three colors are visible.
For instance, if you see orange-yellow-X, then you know which cube it is. The only other time orange and yellow appear on the same cube, they’re rotating in the opposite direction and appear yellow-orange-X.
Wouldn’t it theoretically be possible, in some configurations, to know all six corner pieces + rotation? I think two is a worst-case scenario.
In some configurations (like the apparently-solved case), yes, you can deduce the complete corner state. But there are cases with more than two possible corner states.
Suppose you see
W
R R
R R R
G R R B
G R B
G G B B
G B
W G B W
G B
G B
(that is, all the back corners are white). Then you can identify the hidden corner, but the three white corners can be cyclically permuted, so there are three possible corner states. If instead you see
Y
R R
R R R
G R R B
G R B
G G B B
G B
W G B W
G B
G B
(one yellow and two white corners) then I believe there are five possible permutations of the unknown corners. (The unknown corners are RYW, OGW, OYB, OYW. So listing the possibilities for yellow;white1,white2 we have
RYW;OGW,OYW
OYB;RYW,OGW
OYB;RYW,OYW
OYB;OGW,OYW
OYW;RYW,OGW.)
If the colors you see on the three back corners are instead orange, yellow, white then there are twelve possibilities. This is at least close to the worst case (clearly it can’t be worse than 4!=24 possibilities), and I suspect it actually is the worst case for a more random cube, but I’m not sure of that.
Let’s assume that when edges move, they can only move in threes such that, e.g. edges ABC -> BCA, and the position of the pieces are rotated. Only one rotation manipulation is needed; run the same routine twice and you get the reverse rotation, run it three times and everybody is back to their original position. We further assume all possible manipulation of edges are a basic repetition of this single move over any of the possible cubelet combinations. This alone prevents an outright swap of only two cubelets; it is not possible to get BAC from ABC without using other cubelets and swapping them in the process.
Let’s consider the three matched-color pairs of partially-seen edge cubelets in isolation, based on the conclusion that edges must be moved in threes. This implies either (1) all pairs of these partially-seen edge cubelets are placed properly, or (2) only one of the pairs is properly placed; the other two pairs are swapped. To see this, suppose AB and CD are two pairs. Swap ABC -> BCA to get BC and AD, then swap CAD -> ADC to get BA and DC, the original two pairs with positions switched.
There are also the three unseen edge cubelets. Considered in isolation, and based on the fact that these must be moved three at a time–two alone cannot swap places–there are only three possible configurations for these.
There is one other possibility, namely that two partially-seen edge cubelets are swapped and that two of the three unseen cubelets swap position. This is obviously a case where the two sets of edge cubelets interact.
So, to summarize:
For the case where all partially-seen edge cubelets are properly placed, there are three possible configurations for the unseen edge cubelets: Total of 3 cases.
For each of the three cases where only one pair out of of the partially seen edge cubelets is actually swapped, there are, again, only three possible configurations for the unseen edge cubelets: Total of 9 cases.
For each of the three cases where two of the three pairs of partially-seen edge cubelets are actually swapped, there are, again, only three possible configurations for the unseen edge cubelets: Total of 9 cases.
For the single case where all three pairs are swapped, there are three possible configurations for the unseen edge cubelets: Total of 3 cases.
The total number of possibilities for the position only of the edge cubes is the sum of these, and it totals 24. We must, however, also consider the orientation of the unseen cubes.
As you note, edge cubes must be flipped as a pair; it is impossible to flip a single edge cube. Thus, either all the unseen edges are oriented properly (1 possibility), or only one of the three is (3 possibilities). So, we need to multiply the 24 by these 4 uncorrelated possibilities: 24*4 is 96, which I now believe is the correct answer.
Err…there comes a point where you think you’re solving one problem, but it turns out the request was for something different. I had missed this critical line in the OP:
And I had been assuming the opposite. Please ignore my sheepish grin as I step quitely out of the room…
Oops, these are wrong. The permutations need not be even; so in the first case any of the 6 permutations is possible, and in the second case there are ten permutations (since the two white corners can also be swapped).
Well of course it’s unsolved! Otherwise the answer would be “1”.
But (as I said above) the answer depends on the configuration of the sides you see, so solving the problem in various special cases of high symmetry, like the “apparently-solved” states, is a lot less headache than working out the general case. So I think your posts are to the point of the OP. (And not just because you got the same answer I did, either!)
To have three adjoining, visible sides solved, all the corner cubelets have to be in the correct position, there’s no way around this.
The remaining side cubelets have four sections:
Pairs on the top and each side that can either be right or wrong.
The remaining three that you can not see from the front at all can be right or wrong in six ways.
With this you get:
2 x 2 x 2 x 7 = 56
Subtracting the one possible combo that results in a solved cube you get 55 ways for the backside to be wrong somehow, with only three of those having all the side cubelets wrong.
Don’t worry. Your consolation prize is solving the number of possibilities for the unseen sides of the cube if the Rubik’s cube is on a treadmill.
It is even more embarassing to err in an apology about an err, but at the risk of infinite regression, what I meant was that I was assuming the three visible sides were (or at least appeared to be) solved. But from Omphaloskeptic’s coded diagrams, I see this is a false assumption. My solution, trivial as it is, assumes the three visible sides appear to be solved.
I’m glad I arrived at the same answer as you Omphaloskeptic, a small consolation for what I read now as unjustified hubris. Thx:)
Ack; either that, or it’s trying to guess whether or not to switch to the door Monty Hall didn’t open…