XLV OM - II - Task 4

Each vertex of a cube is assigned the number $ 1 $ or $ -1 $, and each face is assigned the product of the numbers assigned to the vertices of that face. Determine the set of values that the sum of the $ 14 $ numbers assigned to the faces and vertices can take.
Let's consider an arbitrary assignment of the numbers $1$ and $-1$ to the vertices of a cube; denote the sum of these numbers by $S_1$. Assign numbers to the faces according to a given rule; denote the sum of these numbers by $S_2$. The sum of all $14$ considered numbers is $S_1 + S_2$.
If we change the number at one vertex (it doesn't matter which one), this will affect the numbers assigned to three faces. The value of the sum $S_1$ will increase or decrease by $2$. The value of the sum $S_2$ will increase or decrease by $2$ or $6$. The total sum $S = S_1 + S_2$ will increase or decrease by $4$ or $8$, or remain unchanged.
If all vertices are assigned the number $1$, then all faces are also assigned the number $1$, and in this case $S = 8 + 6 = 14$. (This is, of course, the maximum possible value of the sum $S$.) From this configuration, any other can be reached by changing the numbers at the vertices one by one. According to the previous observation, at each step of this procedure, the sum $S$ will change by $\pm 4$, $\pm 8$, or $0$. Therefore, the values taken by the sum $S$ will differ from the initial value $14$ by multiples of the number $4$. The set of possible values is thus a subset of the set $\{14, 10, 6, 2, -2, -6, -10, -14\}$.
The number $-14$ can be immediately eliminated: we would need to have $-1$ at all vertices and all faces, but this is contradictory to the conditions of the problem.
It is also not possible to achieve the value $S = 10$. This would require $12$ ones to appear with a positive sign and two with a negative sign. If all vertices were assigned a positive one, then all faces would also be assigned a positive one. If exactly one vertex were assigned a negative one, then three faces would receive a negative sign. If, finally, exactly two vertices were assigned a negative sign, then some faces would also receive a negative sign. Therefore, the configuration "10 positives, 2 negatives" is not achievable in any way.
The range of possible values for $S$ is thus limited to the set

Of these six values, each can be achieved in various ways. Here are examples of configurations that realize these values:
If all vertices of the cube are assigned the number $+1$, then $S_1 = 8$, $S_2 = 6$, and $S = 14$.
If all vertices of the cube are assigned the number $-1$, then $S_1 = -8$, $S_2 = 6$, and $S = -2$.
If one vertex is assigned the number $+1$ and the rest $-1$, then $S_1 = -6$, $S_2 = 0$, and $S = -6$.
If one vertex is assigned the number $-1$ and the rest $+1$, then $S_1 = 6$, $S_2 = 0$, and $S = 6$.
If two opposite vertices of the cube are assigned the number $+1$ and the rest $-1$, then $S_1 = -4$, $S_2 = -6$, and $S = -10$.
If two opposite vertices of one face are assigned the number $-1$ and the rest of the vertices of the cube are assigned the number $+1$, then $S_1 = 4$, $S_2 = -2$, and $S = 2$.