# Identifying distinct colored cubes from faces of $2 \times 2 \times 2$ arrangement

Note: The answer to Question 1 is "never." But Question 2 remains open!

Consider cubes that can have their faces colored white or red, and let us say that two colorings of a cube are equivalent if one can be rotated to the other, and distinct otherwise.

The number of distinct colorings for cubes that have at least one white face and at least one red face is $8$ (see post script for explanation). Suppose you have an arrangement of those distinct $8$ cubes into a larger, $2 \times 2 \times 2$ cube, and you can see all six of the larger cube's exposed faces.

Question 1: Under what conditions can you tell, from the $2 \times 2 \times 2$ cube's exposed faces, how the original $8$ distinct cubes were arranged? Answer: This is never possible.

For both questions: Although I would be quite pleased to see a "pure mathematical" approach, a [justified, reasoned] brute-force computation would be fine, too.

Question 2: For any given "view" of the $2 \times 2 \times 2$ cube's six faces, there are multiple possible arrangements that could have yielded what one sees. So: What view maximizes the number of possible arrangements, and how many arrangements are there in this maximized scenario?

Post Script. The number of possible cubes with one white face is clearly $1$; for two white faces, either they are adjacent or opposite, which yields $2$ in total; for three white faces, either they all meet at a corner or do not, which yields $2$ in total; for four white faces, it means two red faces, hence $2$ in total; and, for five white faces, it means one red face, hence $1$ in total. So, added up across all colorings of this nature, there are $1 + 2 + 2 + 2 + 1 = 8$ distinct colorings.

• While one way to get the answer to a second question is to edit your original question and then add a bounty, another way is to ask a second question. – Misha Lavrov Dec 22 '17 at 13:19
• @MishaLavrov Yeah; fair enough. – Benjamin Dickman Jun 26 '18 at 17:09