Elijah has a large collection of identical wooden cubes which are plain on $4$ faces and shaded on $2$ faces that share an edge. He glues some cubes together face-to-face. The figure below shows $2$ cubes being glued together, leaving $3$ shaded faces visible. What is the fewest number of cubes that he could glue together to ensure that no shaded faces are visible, no matter how he rotates the figure?

Video solution:

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Written solution:

For no shaded face to be visible, each cube must have both of its shaded faces glued to neighboring cubes, so each cube needs two face-neighbors. With only three cubes, the face-adjacency graph is a path, so an end cube has only one face-neighbor. Four cubes arranged as a $2\times2$ square can each be oriented with its two shaded faces pointing inward, so $4$ cubes suffice.