The Tseng puzzle

This page is under construction. I plan to add here the electronic version of the mathematical analysis of the 2x2x2 and the 3x3x3 cubes described by Daniel Tseng has soon as possible (by now I just have a handwritten version in italian). However I am quite confident on the correctness of the result, at least if I correctly understood the description of the puzzle.

For the impatients, here are the scanned images of my preliminary notes (in italian).

Links

The 2x2x2 Tseng cube

We can prove that by using the three elementary actions (twists, overturns and shifts of a layer) one can reach exactly half of the possible configurations, which are 8! 24^8. More precisely, a configuration g is reachable from the initial configuration (i.e. we can solve the cube starting from g) if and only if the number δ(g) of the visible dark faces is even. Here a dark face of each one of the 8 cubes is any one of the 3 faces that are not visible in the initial configuration.

Note that the shift action as described in this author post can be obtained by using twists and overturns (this is however true only for the 2x2x2 cube), so that we can neglect such elementary actions and consider only the twists and the overturns.

The 3x3x3 Tseng cube

We can prove that by using the three elementary actions (twists, overturns and shifts of a layer) one can reach ALL of the possible configurations, which are 27! 24^27.

Note that we consider to be different two configurations that differ only on the orientation of the central cube; even if it is not visible from the outside. Likewise we consider to be different two configurations where the central cube of one face is rotated of 90 degrees around it's visible face.

The proof will be available soon.

2x2x2 cube: the "only if" proof

Consider all the 6 x 8 = 48 faces of the 8 cubes. We shall call dark a face that is not visible in the initial configuration of the 2x2x2 cube. There are exactly 24 dark faces and 24 non-dark faces.

Given any 2x2x2 configuration g of the 8 cubes, we shall define the function δ(g) as the number of dark faces that are visible. Of course δ of the initial configuration is zero.

To prove the result it is enough to show that any elementary action does not change the parity of δ. This is trivial for all the twist actions, since they do not expose hidden faces and do not hide visible faces, so that δ does not change at all. We only have to consider the overturn actions. Let g' denote the resulting configuration after an overturn action T starting from the configuration g: g' = T(g). Of the 24 visible faces, only 12 are affected by T, and 8 of these (the lateral faces of the affected layer) remain visible after the action. Hence we only have to consider the 4 faces that correspond to the 2x2 face F of the cube affected by T. These will become hidden, and each will be replaced by its opposite face of the cube to which it belongs. Now we observe that each of the 8 cubes have exactly 3 dark faces, located around one of the vertices of the cube, so that the opposite of a dark face is non-dark and viceversa. As a result, if n denotes the number of dark faces among the four faces of F in the configuration g, after the action T they will be replaced by 4 - n dark faces in the configuration g'. The parity of n, and hence of δ, clearly remains unchanged, which concludes the proof.