How to show a Instant Insanity problem has no solution? I was asked to show that there is no solution to the following four cubes instant insanity problem


The game Instant Insanity is played with four cubes. Each face of a cube is colored with one of four colors. The object of the game is to stack the cubes one on top of another so that on each long face of the resulting shape, all four colors show up exactly once. Note that there are four such faces, which we will describe as top, bottom, front, and back. The picture above shows the four cubes placed together in a way that you can see the top and front faces.

One way is to draw out all the possible combinations of the two subgraphs and show that non of the pairs could simultaneously satisfy the requirements of the solution. 
Are there simpler ways of proving a Instant Insanity problem has no solution?
 A: Just like determining lower bounds is generally harder than "finding" upper bounds, so proving no solution for this puzzle is not immediate.  The instance is easier as N = 4, small size, but relatively harder as each color shows up six times. Of course we are assuming we don't have access to a look-up table of all possible obstacles or "blocks" that can occur for puzzles of this size, otherwise we would quickly find a block.  (Had there been a solution then there would be no obstacle and we would have an existence proof for solvable.) 
Suppose one half solution (complete thread) takes 1: b-g. Then it is easy to rule out 2:b-g (no choice for 4), 2: r-y (no choices for 3 and 4).  Therefore if a complete thread exists in this case we must have 2: g-r. And 4: b-r.  And 3: y-y.  Btw, this forms a complete thread.  Now if a complementary thread has 1: R-Y then it must have 2: B-G. And 4: Y-G, leaving no possible choice for cube 3. 
Now if a complementary thread takes 1: G-R then a short check shows 2: R-Y is impossible, leaving 2: B-G.  But it is impossible to complete this thread with what is available left on cubes 3 and 4.  
Thus, the pair 1: r-y MUST be used in any solution.
So we may assume 1: r-y.  Now choosing 2: r-y leaves nothing available for 3.  Choosing 2: g-r also leads, after a couple steps, to a dead end.  Thus we are left with 2: b-g.  There are 4 quick choices to continue this thread to cubes 3 and 4; none of them complete a half solution.  
Obviously this particular instance was constructed so that there is no immediate way to determine the answer, namely, no solution.  The size of the minimal obstacle was 4.  This is typical.
One is tempted to say that puzzles with no solution are harder to determine than ones with solutions.  But supposing someone was utilizing a breadth-first search and the puzzle was a stack of monochromatic cubes so that every arrangement led to a solution.  Then for large instance sizes (say N = 100) the branching would overwhelm the computer's memory.
For a sophisticated depth-first routine to find (all) solutions quickly of puzzles up to size 100, please Google: Chris Reeser.
Finally in Tutte's 1947 paper he says that the chances of hitting on the solution of the particular puzzle he uses as an example is less than one in 40,000.   Due to loops and various symmetries that create multiple solutions the probability is closer to 1/1000.  This of course assumes "blind" stacking, whereas many possibilities would be aborted after 2 or 3 cubes if one was paying attention.  
A: I guess the key is the hamiltonicity of the subraphs...In particular the subgraph have to be connected so you cannot choose edges with loops because the condition "Each vertex must have degree 2" would force you to disconnect your graph. It reduces a lot the possibilies when you try to construct the Hamiltonian subraphs. 
