## Graph Theory Lecture Notes 5

### The Four-Color Theorem

Any map of connected regions can be colored so that no two regions sharing a common boundary (larger than a point) are given different colors with at most four colors.
**History:**

- Francis Guthrie (1850)
- Augustus DeMorgan - Fredrick Guthrie
- Hamilton (1852)
- Cayley (1878) - London Math Society
- Kempe (1879)
- Heawood (1890)
- Appel and Haaken (1976)

Graph theoretical formulation of the problem uses the dual graph obtained from the map by identifying each region with a point and joining these vertices if the regions have a common boundary. These graphs have the property of being planar.
**Reformulated Problem**: Every planar graph can have its vertices colored so that no adjacent vertices have the same color using no more than four colors.

**Kuratowski's Theorem**: A graph is planar if and only if it has no subgraph homeomorphic to K_{5} or K_{3,3}.

**Def**: *k-colorable, chromatic number*.

**König's Theorem** (1936): A graph is 2-colorable iff it has no circuits of odd length.

*Algorithm: Two-Color*

- Initialization: All vertices are colorless. Queue is empty.
- Step 1: Pick a vertex, color it blue, put it in the queue.
- Step 2: Remove the first vertex to be found in the queue, call it y.
- Step 3: Find all the uncolored neighbors of y, color them the opposite color of y, put them in the queue.
- Step 4: If all the vertices are colored stop, else go to Step 2.

This algorithm is an example of a Breadth First Search. It is a good algorithm in that its complexity is O(n+e) = O(n^{2}). Also, note that the algorithm will color any graph with two colors, whether or not this is a 2-coloring depends on the graph.
**Proof of König's Theorem:** Let d(x,v) be the length of the shortest chain between x and v (note it must be simple). If x is colored blue (the first vertex picked in the algorithm), then for each red colored vertex u, we must have d(x,u) odd, while for each blue colored vertex v, we will have d(x,v) even.

Suppose that two neighbors, u and v, get the same color. Then d(x,u) and d(x,v) have the same parity (both odd or both even). The closed chain x - u - v - x is therefore of odd length. If this closed chain is not simple, we can make it simple by removing edges, but this always reduces the number of edges by an even number. When we are finished "pruning" the chain, we will be left with a circuit of odd length.

There are numerous problems that can be answered by appropriately coloring a graph. For instance, scheduling problems, committee assignments, broadcast channel assignments, garbage truck routes.