Coloring a map or graph in such a way that it satisfies the four-color theorem is like solving an ill-conditioned logic problem. Why “ill-conditioned”? A typical logic problem gives you exactly the right amount of information such that if you solve the problem correctly, there is exactly one answer to the problem. When coloring an empty map or graph, the information given in the problem is insufficient, which results in multiple solutions to the problem.
A graph that is constrained such that it is four-colorable only 24 ways can be solved like a well-conditioned logic problem once you set the colors of the vertices of one triangular face. Once these three colors are set, for the most restrictive graphs there is only one way to color the remaining vertices. Logic may be applied to work out the colors of the remaining regions.
If the graph it isn’t highly constrained, meaning that that there are more than 24 ways to color it (or equivalently, after setting the colors of the vertices of one triangular face, there are multiple ways to color the remaining vertices), then a logic table will be ill-conditioned. There won’t be a single solution for coloring the remaining vertices.
Let’s first look at a highly restrictive graph (the corresponding map is shown on the left). If we set B to be color 1, C to be color 2, and D to be color 3, there is only one solution.
For the map and corresponding graph shown on the previous page:
• A connects to B, C, D, F, and G (but not E).
• B connects to A, C, D, E, F, and G.
• C connects to A, B, D, E, and F (but not G).
• D connects to A, B, and C (but not E, F, or G).
• E connects to B, C, and F (but not A, D, or G).
• F connects to A, B, C, E, and G (but not D).
• G connects to A, B, and F (but not C, D, or E).
We can set up an equivalent logic problem based on the sharing between regions. Since B, C, and D are the vertices of a triangular face, these three regions must be three different colors. We may assign colors 1, 2, and 3 to these regions in our logic table.
Now we apply logic:
• A must be color 4 because it can’t be the same as B, C, or D (which are 1, 2, 3).
• F must be color 3 because it can only be the same color as D.
• E must be color 4 because it can’t be the same as B, C, or F (which are 1, 2, 3).
• G must be color 2 because it can’t be the same as A, B, or F (which are 4, 1, 3).
According to our solution to the logic problem, the map and graph can be colored like this:
Setting the problem up with a logic table doesn’t appear to have made the map or graph any easier to color. The logic table merely provides one way to show the logical decision-making involved in the four-coloring.
Now let’s look at a less restrictive graph. If we set A to be color 1, C to be color 2, and E to be color 3 in the graph below, there will be multiple solutions.
For the graph shown on the previous page:
• A connects to B, C, E, and F (but not D).
• B connects to A, C, D, and F (but not E).
• C connects to A, B, D, and E (but not F).
• D connects to B, C, E, and F (but not A).
• E connects to A, C, D, and F (but not B).
• F connects to A, B, D, and E (but not C).
We can set up an equivalent logic problem based on the sharing between regions. Since A, C, and E are the vertices of a triangular face, these three regions must be three different colors. We may assign colors 1, 2, and 3 to these regions in our logic table.
In this example, even after setting the first three colors, the problem remains ill-conditioned since the problem still has multiple solutions. For example, B can’t be the same as A or C, but be could be the same as E. Therefore, B can be color 3 or 4. Similarly, D can be color 1 or 4 and F can be color 2 or 4.
This allows four possible solutions. One is B = 3, D = 1, and F = 2. For the other three solutions, change one of these colors to 4. Note that two regions can’t both be color 4 since regions B, D, and F are all connected to one another.
What would a logic table look like if a solution didn’t exist? One example is to work out a logic table for three-coloring (not four-coloring) of the K4 graph. In the following logic table, region D can’t be colors 1, 2, or 3 because it can’t be the same color as regions A, B, or C, which results in no solution (unless you introduce color 4).
The only way to solve the logic table above is to allow region D to be a fourth color. This proves that K4 is four-colorable, but not three-colorable.
CHAPTER 10 EXERCISES
1. Proceed to complete the logic table for the partially colored graph below.
If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means.
2. Proceed to complete the logic table for the partially colored graph below.
If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means.
3. Proceed to complete the logic table for the partially colored graph below.
If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means.
4. Proceed to complete the logic table for the partially colored graph below.
If you are able to find a unique solution to the logic table, complete the logic table and color the graph according to the completed logic table. If you aren’t able to find a unique solution to the logic table, describe why not, explain what caused the problem, and interpret what this means. Note: This logic table only has three colors. The goal for this logic table is to determine if this graph is or isn’t three-colorable (not whether it is four-colorable).
Challenge problem 1: A K4 subgraph is four-colorable, whereas a K5 subgraph isn’t four-colorable. Is it possible for a graph to contain a subgraph that is a subdivision of K5 (which would make the graph nonplanar according to Kuratowski’s theorem) to be four-colorable? Either provide an example, prove that it isn’t possible, or argue why it would be very difficult to determine. Similarly, is it possible for a graph to contain a subgraph that is a subdivision of K4 to be three-colorable? Either provide an example, prove that it isn’t possible, or argue why it would be very difficult to determine.
Could the answers to these questions help to prove the four-color theorem?
Explain. Are the answers to the above questions relevant to the challenge problem from Chapter 6? Explain.
Note: The answer key doesn’t include answers to the challenge problems.
These problems are intended to encourage you to think about the ideas.
However, you may wish to consider how this problem relates to Chapter 27.