I've learned from How many colors of a Rubik's Cube must be known to locate all? that you need to know a minimum of 17 stickers (spread across all its faces) for a 3x3x3 cube in order to compute its state, but if we just see three full adjacent faces and all the 27 stickers on those faces, can we always compute the cube state considering that the cube is always solvable?
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2Is the cube assumed to be solvable? – Arthur Sep 14 '22 at 11:02
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yes, I'll add this assumption to my description, thanks for pointing it out – Stefan Sep 14 '22 at 11:13
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No. Changing the orientation of any two edge cubies is possible by valid moves. (An edge cubie is a plastic piece with two stickers, always found next to two center stickers between moves.) So is changing the positions of any three edge cubies by an even permutation. Three edge cubies are not visible at all in this view of the cube, so we cannot deduce their positions or orientations.
This also means it's possible to have a cube which appears completely solved from that view and is solvable, but is not actually solved.
(Corner cubies are trickier. It's not always possible to deduce their states, but sometimes, as in the case of a cube appearing solved, we can.)
aschepler
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No. To expand on @aschepler's excellent answer, consider
$$ F'L'(R^2URUR'U'R'U'R'UR')LF $$
(the sequence in the parenthesis is just a standard edge switch).
What you should have at the end of this algorithm is a cube that you can view from one angle that looks completely solved, but it isn't.
timidpueo
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