Instant Insanity Instructor’s GuideMake-it and Take-it Kit for AMTNYS 2006
THE KIT:
This kit contains materials for two Instant Insanity games, a student activity sheet with answer key and this instructor’s guide. This game can be used as an enrichment activity for grades 6-12.
Included Materials:
8 wooden cubes
48 colored stickers in 4 colors
Student activity sheet and answer key
Instructor’s guide
Directions:
For the first set of four cubes, place stickers on the sides of each cube to correspond to the diagrams below (R: red, Y: yellow, B: blue, W: white). The colors are given in the form of a net for the cube, as if the cube is unfolded.
Block 1Block 2Block 3Block 4
RYWW
RW B RB B WY R YR W
YYBW
YWRB
For the second set of cubes, place stickers according to the diagrams below.
Block 1Block 2Block 3Block 4
BWBY
WW R RR B WY Y RR B
YBBW
YYRB
(Make sure to keep these sets separate! Otherwise, there is no guarantee that the set will have a solution! Two Ziploc bags are provided, one for each set.)
The Word documents and .pdf files for this packet and the activity sheet for students can be downloaded in their full 8”x11.5” size from
ACTIVITY:
Topic: Using mathematics (combinations and graph theory) to model a game (Instant Insanity) and develop a strategy to win it.
Objectives:
- Students will explore the Instant Insanity game and discover some of the challenges and constraints.
- As a class, students will use the strategy of working backward from a solution to develop a plan to win the game.
- Students will use the graph theory modelindependently to win Instant Insanity.
Time: The standard activity can be done in a 40 minute period with appropriate scaffolding for your group. With extensions, it can be extended to two 40-50 minute periods, plus continuing investigation by students.
Materials:
- 1 set of 4 blocks for each pair of students
- Worksheet for each student
- Writing utensils (optional: colored pens or pencils so that the edges on the graphs can be color coded)
INSTANT INSANITY:
How to play:
You are given four cubes with sides of four different colors. To win, you must 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.
Goal: Develop a strategy to win the game quickly.
Suggested use of the activity and rationale: (Timing given for a 40-45 minute period)
1. Let students play the game for a few minutes to get some experience with trial and to win. (5 min.)
A few may be able to discover the solution, but most will get close – with only two or three sides of the tower lined up. This hands on experience is intended to give them some intuition for the game as well as give them some sense of how difficult it might be to use trial and error.
2. Give students the worksheet and have them start working on it. (8-10 min.)
You can adjust to fit your students’ backgrounds and abilities and the time constraints. You can also choose to do as much or as little of the worksheet as a class rather than in small groups.
The first question is intended to show them how difficult it is to win the game by chance. Most students – even those that have not studied combinations or probability - can be led through this question, although many might not be able to do it on their own. However, in the interest of time, you may choose to drop this and find another way to convince them how long it would take; for instance, if we assume there is only one solution, and if we try one tower every 15 seconds it could be 41472/4=10368 minutes or 172.8 hours or 7.2 days until we find the solution – more than a week!
The second question is intended to help them think about the constraints of the game and to motivate the solution method. You can demonstrate this by setting up one side of a tower with all four colors and then looking at the other side and asking them if they can keep the first side and change this back.
You can also demonstrate how the other sides can rotate while fixing these opposite sides.
3. Work together as a class to develop the strategy and solve the sample set of cubes. (15-20 min.)
The third question is meant to help them develop a strategy to win the game while teaching mathematical strategies of working backward from a solution and using a model to think about the game. You can choose to present as little or as much background about vertex-edge graphs as you wish (more information for you below!) Make the first graphs (I suggest using different colors to represent different cubes!) and then allowing students to share any patterns they notice.
After noting the three key points, we need to turn the idea around and work forward from the four blocks. The big conceptual key is that we can’t start by deciding which opposite pairs of sides go on which graph – so we put all the opposite pairs on ONE graph and then have to find subsets of these edges that have the three required properties. At this point in the strategy, there may actually be more than one solution! Most students will probably find it easier to be led through this process the first time.
These subsets of edges are called subgraphs. A good strategy for finding subgraphs is to pick one cube and look at its edges one at a time. We know we have to have an edge from each cube, so it doesn’t matter which cube to pick. Picking one cube allows us to limit the choices. For example, for the first sample set of cubes, if we pick the fourth cube and the W-W edge which represents two whites on opposite sides of the cube, we know that for this edge to be used in a subgraph, the other three edges must form a triangle with the other three vertices. This is impossible since there is no edge between the B and Y vertices – so we know that we can’t use this W-W edge from the fourth cube at all! We then go on to the other edges from the fourth cube and try to find solutions. If we consider the W-Y edge from the fourth cube, either we have to have all four edges form a square (W-Y, Y-B, B-R, R-W) or an hourglass (W-Y, Y-R, R-B, B-W). Again, the first possibility is impossible since there are no B-Y edges! Looking at the diagonal edges, we see we can take either diagonal from cube 2 or 3, so we need to get an edge from cube 1 between R and B. Then we have two choices for a subgraph, depending on whether we take the Y-R edge from cube 2 or cube 3. We would then go on to try to find another subgraph using the last edge from cube 4.
4. Allow students to practice the strategy with a different set of Instant Insanity cubes in groups. (5-10 min.)
Students can be given a different set of cubes (for example, the second pattern of cubes in the kit) to work with or they can be asked to work more abstractly, just using the nets. There is also an online applet for Instant Insanity which will give multiple examples for them to try. This can be found at To make up your own cubes, I suggest starting by stacking four cubes, making sure each side of the tower has a cube with each color, and then randomly coloring the faces that were on the top and bottom. You can then check your cubes using the solution strategy!
Options for cost-effective cubes:
- Make nets of cubes for students and have them cut them out and fold them up to use.
- Use sugar cubes that are colored on the sides with markers (a set can be stored in an envelope)
- Buy wooden cubes in bulk; I bought from Woodworks, Ltd. at
SOLUTION TO THE SECOND SAMPLE SET OF CUBES:
The graph for this set of cubes is as follows:
Block 1Block 2Block 3Block 4R B
BWBY
WW R RR B WY Y RR B
YBBW
YYRB
W Y
One pair of winning subgraphs:
RBRB
WYWY
EXTENSIONS:
1. 5-cube Instant Insanity
Instant Insanity can also be played with five colors and five blocks. Here are the colored nets for a set of 5 cubes if you wish to challenge students to solve this:
Block 1Block 2Block 3Block 4Block 5
RYWBR
GWYRGWBGYYGWWYR
WBRBB
BGYRG
The solution to 5-cube Instant Insanity is almost exactly the same as the answer to 4-cube Instant Insanity. The main difference is that you will have five vertices in your graph rather than four, since there are now five colors. We still need to find two subgraphs (one for the front-back, one for the left-right) where
1) each subgraph has five edges,and each edge is in only one subgraph.
2) each color vertex in each subgraph has two edges.
3) there is an edge in each subgraph from each cube.
SOLUTION: R
graph of opposite sides:
cube 1:Y B
cube 2:
cube 3:
cube 4:
cube 5:
W G
One pair of winning subgraphs:
RR
YB YB
WG WG
Here is an extra challenge (even for teachers!):
Moshe’s Insanity: You have 8 blocks with 6 different colors of faces. You want to form them into a 2x2x2 cube so that the cube’s faces are 6 different solid colors. Can you use methods you learned from Instant Insanity to find a winning strategy? Here is a sample set of cubes with two very similar solutions:
1:2:3:4:
OOOO
YR P BB Y PP G PY G
GGYR
BRRB
5:6:7:8:
GYRR
BY R BG O YO G OY G
POOP
ORRB
This is MUCH more complicated. Of course, there are also many more combinations of blocks! The key here is to note the oriented corners of the smaller blocks. While the constraint for Instant Insanity was opposite pairs of sides, the oriented triples of colors (three colors in order around a vertex) at the corners will be the constraint for this problem. These corners will be seen in the larger block and they must be set up so that a whole large block can have these oriented triples. (If you have R-Y-B counterclockwise around a corner vertex of the final cube, you need to find a small cube that has this color combination in the same orientation – having it clockwise doesn’t work!) However, to begin with we don’t know exactly which corners the final cube will have. But we can list all the oriented triples that occur on the smaller cubes and most of the possible final cubes will be ruled out because some of the oriented triples do not occur on the smaller blocks. (There are 6·5·4=120 possible oriented triples and at most 82=64 on the cubes!)
At this point, we can choose to model this by a series of graphs, one for each possible final cube. We make a graph that has a vertex for the oriented triple at each corner of the final cube and a vertex for each of the eight small cubes. We usually draw these as two horizontal lines of eight vertices, one line above the other. We then draw edges between an oriented triple vertex and a small cube vertex if the small cube has that oriented triple. (Since all edges are drawn between the two sets of vertices and never connect to small cubes or two oriented triples, this is called a bipartite graph, since it can be partitioned into two sets.)
Here is a sample graph:
If the final cube has a net like that below, its (counterclockwise) oriented triples are G-R-B,
R-Y-B, P-R-G, Y-R-P, O-P-G, B-O-G, Y-P-O, Y-O-B
B
YR G Note that R-Y-B, Y-B-R and B-R-Y are the same oriented triple since it
Pis the counterclockwise order that matters. So we will rewrite each of
Othe triples so it starts with the alphabetically first letter, and label a vertex
for each. The bottom set of vertices is labeled by cube.
B-G-R B-R-Y B-O-G B-Y-O G-P-R G-O-P O-Y-P P-Y-R
1 2 3 4 5 6 7 8
Now it is a matter of finding if there is a way to match up every cube with one oriented triple so each cube gets used and every oriented triple is found. In the graph, this means finding a set of eight edges that is incident on all sixteen vertices – that is, no vertex has two edges from the set that are incident to it. This is called a perfect matching. If we can find a perfect matching, and an edge connects a small cube to a certain oriented triple, then this small cube needs to be placed in the larger cube so that the oriented triple is the part of the small cube that would show in the final cube. Clearly, the graph above doesn’t have a perfect matching – there are a number of vertices that have no edges incident to them. For example, the oriented triples on cube 1 do not at all match the oriented triples of the final cube, so there are no edges to vertex 1 and there would be no place cube 1 could be put so it could be a part of the final cube. This is an example of a final cube we could eliminate very quickly. Other graphs are harder to determine if there are perfect matchings. Hall’s theorem says that there is a perfect matching in a bipartite graph if and only if for every set of vertices we pick in either the top (or bottom) set of the bipartite graph, there are at least as many vertices in the other set that are adjacent to our first set. For example, in the graph above, if we pick the set of three vertices 2, 3, and 7, the vertices that are adjacent are B-G-R and P-Y-R – only two vertices, so we know there isn’t a matching.
BACKGROUND:
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 edges are 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:
B
3 4
A2 C
1 7 6 5
ED
An edge which has both endpoints at the same vertex, such as the edge labeled 4, is called a loop. We say that an edge is incident to a vertex if the vertex is one of its endpoints. Therefore we can say that edge 3 is incident to vertex B and vertex C. Two vertices are adjacent if they are connected by an edge. Since edge 6 connects C and E, these two vertices are adjacent. The degree of a vertex is the number of edges that are incident to the vertex, counting twice if an edge is a loop. Therefore, vertex A has degree two while vertex C has degree five.
Subgraphsare any subset of vertices and edges of the original graph. For Instant Insanity, we want special subgraphs that include all the vertices and where each vertex has degree two. However, in general, subgraphs do not have to include all vertices or have the same degree at each vertex.
Graphs can be used to model many applications. They are used widely in computer science to model computer networks (modeling a computer with a vertex and a network connection with an edge) and to plan algorithms to program computers. Graphs are frequently encountered as a way to model the “handshake problem”: if there are n people in the room, and each shakes everyone else’s hand, how many handshakes take place? This can be modeled with n vertices and an edge between every pair of vertices (no loops here!). The resulting number of edges is the number of handshakes, nC2. This graph is called a complete graph since every possible (non-loop) edge is included. Three other famous problems involving graphs are the Königsberg bridge problem, the traveling salesman problem, and the map-coloring problem.
The Königsberg bridge problem was posed by Euler, asking if there is a way to walk across all the bridges in 18th century Königsberg exactly once and to end up where you started. When the land masses are modeled with vertices and the bridges with edges, this is reduced to the question of whether we can travel along each edge of the graph exactly once and end up where we started. It was impossible in Königsberg, but we can ask the same question for any graph. There is a simple constraint: we can walk each edge exactly once and end up where we started if and only if each vertex has even degree. Such a walk is called an Eulerian circuit. This could also be used to model the solution to an optimum route that a postal worker could follow in order to travel every road and therefore reach every mailbox.
The traveling salesman problem asks what the shortest path is for a salesman to visit a number of cities and return home. This is modeled by a graph where the vertices represent cities and the edges represent roads (orairline flights) between the cities. A key addition here is that the edges are assigned a length or weight so we can tell the distance the salesman would travel on that road. It turns out that this is a very difficult problem to solve quickly; as the number of cities increases, the best algorithm that is known increases in time exponentially! (This is an example of an NP-Complete problem, a special class of problems in computer science.) If we once again disregard the lengths of the edges, any path that would take us to each vertex exactly once and bring us back to where we started is a Hamiltonian circuit. Note that in this case, we will rarely travel along all the edges as we did in an Eulerian circuit! If we can visit each vertex without repeating any, but can’t return to our starting place, this is called a Hamiltonian path.
The map-coloring problem asks what the smallest number of colors is that can be used to color a map so that no neighboring countries have the same color. (It is stipulated that each country is connected – it can’t be split into two pieces that are required to have the same color). This is sometimes also called the four-color problem since it has now been proved that four colors is the required number. This proof was done in 1976 by Appel and Haken and is still considered somewhat controversial because it required a computer to check a large number of cases. This problem can be modeled by a graph by representing each country with a vertex and drawing an edge between two vertices if the two countries share a border. Then we need to find the smallest number of colors to color the vertices so that no two vertices that are adjacent have the same color. For a specific graph, this smallest number is called the chromatic number. We can draw graphs with chromatic number as large as we want (for example, we the complete graph on n vertices has chromatic number n), but graphs that come from maps have special properties – they can be drawn so none of the edges cross. Such graphs are called planar graphs. Because of this special property, any graph that comes from a map has chromatic number at most four!