2. What is it?
The instant insanity game consists of four cubes which
are colored differently on each face with the colors red,
white, blue, and green. The object of the game is given
four of these blocks the player is to figure out how to
stack the four blocks on top of one another so as to
obtain four different colored faces on each side of the
stack. There are many different games and many
different solutions.
The problem with this game is in using four blocks
which can each be placed in 24 different positions in
the stack the possibility of stacks is:
242 = 331,776 possible stacks
That is a lot of stacks!
3. How do you solve it?
One way is to play around with the cubes for hours until
you figure it out. However, there is a math approach
using graphs and graph theory:
One way to look at this problem is by the decomposition
principle:
Start with just looking at the left side and right side of
the stack
Then rotate the blocks to align the back and front to
solve the puzzle
Sounds easy right?…Let’s make a graph
4. Recall: The Multigraph
A multigraph is a graph in which it is permitted for
several edges to join the same pair of vertices. Loop
edges, where an edge connects in a loop back to the
same vertex, is also allowed.
A
B
C
D
6. Now for this puzzle we will create a multigraph. There will be
four vertices each representing the four colors of the faces.
Connect the vertices with edges when they are opposite from
each other. Label the cubes.
1
1
1
2
2
2
3
3
3
4
4
4
R
G
B
R
Cube 1
7. To Solve:
Find a labeled factor.
A labeled factor is a simple disjoint circuit or factor of the
graph where each term (cube number) appears once.
1 2
3
4
1
2
3
4
1
2
3
4
8. Make a Stack
Now arrange the blocks according to the subgraphs we
constructed. The arrow points to the block face that faces right.
1 2
3
4
1
2
3
4
9. Insane Yet?
Graph-theoretic Formulation of Instant Insanity:
Find two edge-disjoint labeled factors in the graph of the
Instant Insanity puzzle, one for left-right sides and one for
front-back sides
Use the clockwise traversal procedure to determine the
left-right and front-back arrangements of each cube.
10. In the previous labeled factor graphs no two were
disjoint because they all contained the same 1 edge.
1 2
3
4
1
2
3
4
1
2
3
4
1
2
3 4 Another
factor
