Instant Insanity – answer key
How to play:
You are given four cubes whose sides are colored with four different colors. Can you find a way to stack the cubes in a tower so that from whichever side you look at the tower, you see all four colors?
Our goal today is to find a winning strategy for this game so you can solve any stack in about 5 minutes!
How to win:
1) How many different ways are there to set up the cubes in a stack? (Note that moving a cube from the top to the bottom or the bottom to the middle while maintaining which colors are on which side results in a stack that is essentially the same since ordering of the colors on the stack does not matter.)
3(24)3=41472. First pick a cube and place it on the table – by the note, it doesn’t matter which cube. There are three essentially different ways to do this – a question of which pair of faces will not get seen on the side (which of the faces is on the top or the bottom of the cube only changes whether the stack has to get turned over once it is formed). For each of the next three cubes, there will be the same number of ways to place each of them. For each cube, we must again choose which faces get hidden, but now which face is on the top and which face is on the bottom will make a difference since it will change whether the faces around the sides come clockwise or counterclockwise. Therefore there are six ways to choose which faces show on the sides and in which order they come – one for each face that we can put down on the previous cube. After this, we can still rotate the cube into four different positions. Therefore, there are 6x4=24 ways to place each of the second through fourth cubes. As a result, the total number of possibilities is 3(24)3=41472. This is WAY too many possibilities to want to try!
This can be scaffolded by breaking it up into the following pieces (and even further if necessary):
a) How many essentially different ways can the first cube be placed on the table?
b) How many different ways can you put down the second cube? The third cube? The fourth?
c) We can get the total number of ways that we can put together the cube by multiplying together the number of ways we can place each cube. (This is called the Fundamental Counting Principle.) What is the total number of ways to put together a stack of four cubes?
2) What seems to be the constraint in getting colors to line up? If you have one side of the tower in all different colors, which colors in the tower can’t you change anymore?
The colors on the opposite side. Everything else we can rotate around these faces.
We will use a tool called a vertex-edge graph (or simply, a graph) to model what is happening with the colors on opposite sides of this stack. Vertices are drawn as points and edgesare curves that connect two vertices (both vertices can be the same!). Even if we draw two edges so they cross, we don’t think of that as creating a new vertex. The following is a graph with 5 vertices and 7 edges:
3) Consider a stack that worked:
You have a stack of blocks with the following alignment of colors:
BlockLeftFrontRightBack
1RWBR
2YRYB
3BYRW
4WBWY
Draw two graphs, one for each of the front-back and left-right pairs. For each, the vertices are given labeled by colors. On the front-back graph, draw an edge between two colors if one of the blocks has those colors on the front and back faces. Label edges by the number of the block the pair is from. Draw a similar graph for the left-right pairs.
Front-BackLeft-Right
RBRB
WYWY
What do you notice about the graphs? What properties do they both have? Are the propertiesenough to guarantee that the stack will work? Why or why not?
Both graphs have one edge per cube and two edges per color(vertex).
This will guarantee that the stack will work, since we can follow the edges around and make the first vertex of each edge the front and the second the back (or left and right respectively) so that each side of the stack has all different colors.
So how can we use what we learned to FIND a solution?
Above we noticed that it is opposite pairs of faces that matter!
We also noticed that if we draw edges on a graph for these pairs, we need to find two sets of edges(called subgraphs) where:
1) each subgraph has four edges,and each edge is in only one subgraph.
2) each color vertex in each subgraph has twoedges.
3) there is an edge in each subgraph from each cube.
Try drawing a graph for the set of blocks that you have! For convenience, there are spaces on the paper to put the blocks so you remember which is which.
Example:
Block 1Block 2Block 3Block 4Graph:
RYWWRB
RWBRBBWYRYRW
YYBW
YWRB
WY
To win, we need to find two subgraphs that have the properties we listed above. Try to find two such subgraphs and draw them below!
Subgraphs:(you can also switch which cubes the R-B and W-Y edges of the first graph come from, so there are two solutions.)
RB RB
WYWY
Now you can put the cubes together to win! Use the first subgraph to know which pairs of edges from which cubes to line up on the front and back. Then use the second subgraph to know which of the remaining two pairs on each cube goes to the left and right.
Try using this same method on another set of cubes!
Challenge:
What if you have 5 cubes with 5 colors? Can you get a method of solution for such a set?
The method of solution is very similar…make a graph with opposite faces as edges, but now you have 5 colors, so there are 5 vertices. You still want to find two subgraphs using 10 total edges, now, where each vertex has degree 2 and there is an edge from each block.