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A "Sudoku cube" is a 3x3x3 uncoloured Rubik's cube. In the solved state, each face has the digits 1 through 9 arranged in ascending rows from top to bottom, and all of the digits on a given face have the same orientation. When designing such a cube, each of the 6 faces has 4 possible orientations in a fixed frame of reference. Assume that cube designs that can be transformed into one another by rotation of the entire cube are equivalent.

The orientations of the faces affects the number of possible solutions to the cube, if each cuboid is taken as distinct. Take the following cube, where the orientation of a face is given as the direction of the top of the digits on that face:

  • Top face—backward.

  • Back face—downward.

  • Bottom face—frontward.

  • Front face—upward.

There are four possible positions that the left and right faces can each take while preserving the solved state, yielding 16 obvious solutions and probably many more.

What set of orientations for all of the cube's faces yields the lowest number of valid solutions, and what is that lower bound?

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On one hand, this is an interesting question.

However, it's also ridiculous. The amount of computation necessary to mostly solve the number of moves needed for Rubik's Cube was gigantic. The renderfarm for Spiderman 3, a system worth many millions of dollars, wasn't enough. It took the entire spare capacity of the Google network several weeks to solve it. See for details.

The amount of programming / computational power necessary to answer your question make it unlikely that answers will arrive anytime soon.

On the other hand, parity restrictions for Rubik's are easy to calculate. What are the odds that a given randomly assembled cube is unsolvable? I believe it's 1/12 for Rubik. That's to give a very specific solution. A Sudoku cube can have multiple solutions if the cubes are put together freely.

Despite all the work on it, there are still 43,252,003,274,489,855,999 Rubik's positions considered unsolved.

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+1 for giving me some perspective. Intuitively I thought this problem would be easier than most Rubik's problems, but I just wasn't sure how to go about it. – Jon Purdy Jun 9 '11 at 22:18

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It might be helpful to Jon if you expanded on your answer; Silas Pike has already mentioned the correct number. – Zev Chonoles Sep 14 '11 at 1:03
Welcome to SE. Please note that leaving very short and uninformative answers is not very useful. These are answers that are likely to be ignored as someone else is likely to better explain a solution. – Ittay Weiss Nov 12 '12 at 4:46

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