The Instant Insanity puzzle is played with four cubes. Each face of the cube is colored with one of the four colors: red (R), white (W), blue (B) and green (G).

The Instant Insanity puzzle is played with four cubes. Each face of the cube is colored with one of the four colors: red (R), white (W), blue (B) and green (G).

Note: All front faces are WHITE

The objective of the puzzle 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.

Determining the Instant Insanity solution is a very difficult task.

Determining how to arrange these four cubes in a stack that is an Instant Insanity solution is a very difficult task. The problem is that all the cubes are not the same.

Determining how to arrange these four cubes in a stack that is an Instant Insanity solution is a very difficult task. The problem is that all the cubes are not the same.

Observe that there are 24 symmetries of the cube, thus in doing this puzzle with four blocks which can each be placed in 24 different positions; that is:

244 = 331,776 possible stacks.

Wow!!

One way of solving puzzle is to play around with these four cubes until you can figure out the solution, however I promise you that this will drive you insane.

One way of solving puzzle is to play around with these four cubes until you can figure out the solution, however I promise you that this will drive you insane.

The other way to look at this problem is by the decomposition principle:

Pick one pair of opposite faces on each cube for left and right sides of the stack so that these two sides of the stack will have one face of each color.

Then pick a different pair of opposite faces on each cube for the front and back sides of the stack so that these two sides will have one face of each.

A multigraph is a generalized graph in which multiple edges are allowed, that is two or more edges can join the same two vertices; and loops are allowed, that is edges of the form (a , a).

Make one vertex for each of the four colors.

Make one vertex for each of the four colors.

For each pair of the opposite faces on cube i, create an edge with label i joining the two vertices representing the colors of these two opposite faces.

For opposite faces: l1=blue, r1=white on cube 1, draw edge labeled 1 between blue and white; for f1=red,b1=red, draw a loop labeled 1 at red; and t1=green, u1=blue, drew edge labeled 1 between green and blue.

For opposite faces: l2=white, r2=blue on cube 2, draw an edge labeled 2 between white and blue; for f2=white,b2=white, draw a loop labeled 2 at white; and for t2=white, u2=red, drew an edge labeled 2 between whiteand red.

For opposite faces: l2=white, r2=blue on cube 2, draw an edge labeled 2 between white and blue; for f2=white,b2=white, draw a loop labeled 2 at white; and for t2=white, u2=red, drew an edge labeled 2 between whiteand red.

For opposite faces: l3=blue, r3=green on cube 3, draw an edge labeled 3 between blue and green; for f3=white,b3=red, draw an edge labeled 3 between white and red; and for t3=red, u3=red, drew a loop labeled 3 at red.

For opposite faces: l3=blue, r3=green on cube 3, draw an edge labeled 3 between blue and green; for f3=white,b3=red, draw an edge labeled 3 between white and red; and for t3=red, u3=red, drew a loop labeled 3 at red.

For opposite faces: l4=green, r4=red on cube 4, draw an edge labeled 4 between green and red; for f4=white,b4=green, draw an edge labeled 4 between white and green; and for t4=blue, u4=green, drew an edge labeled 4 between blue andgreen.

For opposite faces: l4=green, r4=red on cube 4, draw an edge labeled 4 between green and red; for f4=white,b4=green, draw an edge labeled 4 between white and green; and for t4=blue, u4=green, drew an edge labeled 4 between blue andgreen.

Now we can construct a single multiple graph with 4 vertices and all the 12 edges associated with the 4 cubes.

Now we can construct a single multiple graph with 4 vertices and all the 12 edges associated with the 4 cubes.

Now you restate decomposition principle in terms of subgraphs.

Now you restate decomposition principle in terms of subgraphs.