12.10 Trees

What Is A Tree?

Whether we are talking about a family tree or a tree in a forest, none of the branches ever loops back around and rejoins the trunk. This means that a tree has no cyclic subgraphs, or is acyclic. A tree also has only one component. So, a tree is a connected acyclic graph. Here are some graphs that have the same characteristic. Each of the graphs in Figure 12.204 is a tree.

Figure 12.204 Graphs T, P, and S

Let’s practice determining whether a graph is a tree. To do this, check if a graph is connected and has no cycles.

Example 12.46

Identifying Trees

Identify any trees in Figure 12.205. If a graph is not a tree, explain how you know.

Figure 12.205 Graphs M, N, and P

Solution

• Graph M is not a tree because it contains the cycle (b, c, f).
• Graph N is not a tree because it is not connected. It has two components, one with vertices h, i, j, and another with vertices k, l, m.
• Graph P is a tree. It has no cycles and it is connected.

Your Turn 12.46

1.

There are some configurations that are commonly used when setting up computer networks. Several of them are shown in the given figure. Which of the configurations in the figure appear to have the characteristics of a tree graph? If a configuration does not appear to have the characteristics of a tree graph, explain how you know.

Common Network Configurations

Types of Trees

Mathematicians have had a lot of fun naming graphs that are trees or that contain trees. For example, the graph in Figure 12.206 is not a tree, but it contains two components, one containing vertices a through d, and the other containing vertices e through g, each of which would be a tree on its own. This type of structure is called a forest. There are also interesting names for trees with certain characteristics.

• A path graph or linear graph is a tree graph that has exactly two vertices of degree 1 such that the only other vertices form a single path between them, which means that it can be drawn as a straight line.
• A star tree is a tree that has exactly one vertex of degree greater than 1 called a root, and all other vertices are adjacent to it.
• A starlike tree is a tree that has a single root and several paths attached to it.
• A caterpillar tree is a tree that has a central path that can have vertices of any degree, with each vertex not on the central path being adjacent to a vertex on the central path and having a degree of one.
• A lobster tree is a tree that has a central path that can have vertices of any degree, with paths consisting of either one or two edges attached to the central path.

Examples of each of these types of structures are given in Figure 12.207.

Figure 12.206 Forest Graph F

Figure 12.207 Six Types of Trees

Example 12.47

Identifying Types of Trees

Each graph in Figure 12.208 is one of the special types of trees we have been discussing. Identify the type of tree.

Figure 12.208 Graphs U and V

Solution

Graph U has a central path a → b → d → f → i → l → o → q. Each vertex that is not on the path has degree 1 and is adjacent to a vertex that is on the path. So, U is a caterpillar tree.

Graph V is a path graph because it is a single path connecting exactly two vertices of degree one, r → s → u → v → w.

Characteristics of Trees

As we study trees, it is helpful to be familiar with some of their characteristics. For example, if you add an edge to a tree graph between any two existing vertices, you will create a cycle, and the resulting graph is no longer a tree. Some examples are shown in Figure 12.209. Adding edge bj to Graph T creates cycle (b, c, i, j). Adding edge rt to Graph P creates cycle (r, s, t). Adding edge tv to Graph S creates cycle (t, u, v).

Figure 12.209 Adding Edges to Trees

It is also true that removing an edge from a tree graph will increase the number of components and the graph will no longer be connected. In fact, you can see in Figure 12.210 that removing one or more edges can create a forest. Removing edge qr from Graph P creates a graph with two components, one with vertices o, p and q, and the other with vertices r, s, and t. Removing edge uw from Graph S creates two components, one with just vertex w and the other with the rest of the vertices. When two edges were removed from Graph T, edge bf and edge cd, creates a graph with three components as shown in Figure 12.210.

Figure 12.210 Removing Edges from Trees

A very useful characteristic of tree graphs is that the number of edges is always one less than the number of vertices. In fact, any connected graph in which the number of edges is one less than the number of vertices is guaranteed to be a tree. Some examples are given in Figure 12.211.

Figure 12.211 Number of Vertices and Edges in Trees vs. Other Graphs

Example 12.48

Exploring Characteristics of Trees

Use Graphs I and J in Figure 12.212 to answer each question.

Figure 12.212 Graphs I and J

1. Which vertices are in each of the components that remain when edge be is removed from Graph I?
2. Determine the number of edges and the number of vertices in Graph J. Explain how this confirms that Graph J is a tree.
3. What kind of cycle is created if edge im is added to Graph J?

Solution

1. When edge be is removed, there are two components that remain. One component includes vertices a, b, and c. The other component includes vertices d, e, and f.
2. There are seven vertices and six edges in Graph J. This confirms that Graph J is a tree because the number of edges is one less than the number of vertices.
3. The pentagon (i, h, j, l, m) is created when edge im is added to Graph J.

Spanning Trees

Suppose that you planned to set up your own computer network with four devices. One option is to use a “mesh topology” like the one in Figure 12.213, in which each device is connected directly to every other device in the network.

Figure 12.213 Common Network Configurations

The mesh topology for four devices could be represented by the complete Graph A₁ in Figure 12.214 where the vertices represent the devices, and the edges represent network connections. However, the devices could be networked using fewer connections. Graphs A₂, A_3, and A₄ of Figure 12.214 show configurations in which three of the six edges have been removed. Each of the Graphs A₂, A₃ and A₄ in Figure 12.214 is a tree because it is connected and contains no cycles. Since Graphs A₂, A₃ and A₄ are also subgraphs of Graph A₁ that include every vertex of the original graph, they are also known as spanning trees.

Figure 12.214 Network Configurations for Four Devices

By definition, spanning trees must span the whole graph by visiting all the vertices. Since spanning trees are subgraphs, they may only have edges between vertices that were adjacent in the original graph. Since spanning trees are trees, they are connected and they are acyclic. So, when deciding whether a graph is a spanning tree, check the following characteristics:

• All vertices are included.
• No vertices are adjacent that were not adjacent in the original graph.
• The graph is connected.
• There are no cycles.

Example 12.49

Identifying Spanning Trees

Use Figure 12.215 to determine which of graphs M₁, M₂, M₃, and M₄, are spanning trees of Q.

Figure 12.215 Graphs Q, M₁, M₂, M_3, and M₄

Solution

1. Graph M₁ is not a spanning tree of Graph Q because it has a cycle (c, d, f, e).
2. Graph M₂ is a spanning tree of Graph Q because it has all the original vertices, no vertices are adjacent in M₂ that weren’t adjacent in Graph Q, Graph M₂ is connected and it contains no cycles.
3. Graph M₃ is not a spanning tree of Graph Q because vertices a and f are adjacent in Graph M₃ but not in Graph Q.
4. Graph M₄ is not a spanning tree of Graph Q because it is not connected.

So, only graph M₂ is a spanning tree of Graph Q.

Your Turn 12.49

Use the given figure for the following exercises.

1.

Since sq is not an edge in Graph H, Graph N₁ cannot be a spanning tree of H.

1. True
2. False

2.

Graph N₂ is a spanning tree of Graph H.

1. True
2. False

3.

Graph N₃ is a spanning tree of Graph H.

1. True
2. False

4.

Since there is no path between p and t in Graph N₄, it cannot be a spanning tree of any graph.

1. True
2. False

Constructing a Spanning Tree Using Paths

Suppose that you wanted to find a spanning tree within a graph. One approach is to find paths within the graph. You can start at any vertex, go any direction, and create a path through the graph stopping only when you can’t continue without backtracking as shown in Figure 12.216.

Figure 12.216 First Phase to Construct a Spanning Tree

Once you have stopped, pick a vertex along the path you drew as a starting point for another path. Make sure to visit only vertices you have not visited before as shown in Figure 12.217.

Figure 12.217 Intermediate Phase to Construct a Spanning Tree

Repeat this process until all vertices have been visited as shown in Figure 12.218.

Figure 12.218 Final Phase to Construct a Spanning Tree

The end result is a tree that spans the entire graph as shown in Figure 12.219.

Figure 12.219 The Resulting Spanning Tree

Notice that this subgraph is a tree because it is connected and acyclic. It also visits every vertex of the original graph, so it is a spanning tree. However, it is not the only spanning tree for this graph. By making different turns, we could create any number of distinct spanning trees.

Example 12.50

Constructing Spanning Trees

Construct two distinct spanning trees for the graph in Figure 12.220.

Figure 12.220 Graph L

Solution

Two possible solutions are given in Figure 12.221 and Figure 12.222.

Figure 12.221 First Spanning Tree for Graph L

Figure 12.222 Second Spanning Tree for Graph L

Your Turn 12.50

1.

Construct three distinct spanning trees for Graph J.

Graph J

Revealing Spanning Trees

Another approach to finding a spanning tree in a connected graph involves removing unwanted edges to reveal a spanning tree. Consider Graph D in Figure 12.223.

Figure 12.223 Graph D

Graph D has 10 vertices. A spanning tree of Graph D must have 9 edges, because the number of edges is one less than the number of vertices in any tree. Graph D has 13 edges so 4 need to be removed. To determine which 4 edges to remove, remember that trees do not have cycles. There are four triangles in Graph D that we need to break up. We can accomplish this by removing 1 edge from each of the triangles. There are many ways this can be done. Two of these ways are shown in Figure 12.224.

Figure 12.224 Removing Four Edges from Graph D

Example 12.51

Removing Edges to Find Spanning Trees

Use the graph in Figure 12.225 to answer each question.

Figure 12.225 Graph V

1. Determine the number of edges that must be removed to reveal a spanning tree.
2. Name all the undirected cycles in Graph V.
3. Find two distinct spanning trees of Graph V.

Solution

1. Graph V has nine vertices so a spanning tree for the graph must have 8 edges. Since Graph V has 11 edges, 3 edges must be removed to reveal a spanning tree.
2. (a, c, d), (a, c, f), (a, d, c, f), and (b, e, h, i, g)
3. To find the first spanning tree, remove edge ac, which will break up both of the triangles, remove edge cf , which will break up the quadrilateral, and remove be, which will break up the pentagon, to give us the spanning tree shown in Figure 12.226.

   Figure 12.226 Spanning Tree Formed Removing ac, cf, and be

   To find another spanning tree, remove ad, which will break up (a, c, d) and (a, d, c, f), remove af to break up (a, c, f), and remove hi to break up (b, e, h, i, g). This will give us the spanning tree in Figure 12.227.

   Figure 12.227 Spanning Tree Formed Removing ad, af, and hi

Who Knew?

Chains of Affection

Here is a strange question to ask in a math class: Have you ever dated your ex’s new partner’s ex? Research suggests that your answer is probably no. When researchers Peter S. Bearman, James Moody, and Katherine Stovel attempted to compare the structure of heterosexual romantic networks at a typical midwestern high school to simulated networks, they found something surprising. The actual social networks were more like spanning trees than other possible models because there were very few short cycles. In particular, there were almost no four-cycles.

Figure 12.228 Chains of Affection

“…the prohibition against dating (from a female perspective) one’s old boyfriend’s current girlfriend’s old boyfriend – accounts for the structure of the romantic network at [the highschool].”

In their article “Chains of Affection: The Structure of Adolescent Romantic and Sexual Networks,” the researchers went on to explain the implications for the transmission of sexually transmitted diseases. In particular, social structures based on tree graphs are less dense and more likely to fragment. This information can impact social policies on disease prevention. (Peter S. Bearman, James Moody, and Katherine Stovel, “Chains of Affection: The Structure of Adolescent Romantic and Sexual Networks,” American Journal of Sociology Volume 110, Number 1, pp. 44-91, 2004)

Kruskal’s Algorithm

In many applications of spanning trees, the graphs are weighted and we want to find the spanning tree of least possible weight. For example, the graph might represent a computer network, and the weights might represent the cost involved in connecting two devices. So, finding a spanning tree with the lowest possible total weight, or minimum spanning tree, means saving money! The method that we will use to find a minimum spanning tree of a weighted graph is called Kruskal’s algorithm. The steps for Kruskal’s algorithm are:

Step 1: Choose any edge with the minimum weight of all edges.

Step 2: Choose another edge of minimum weight from the remaining edges. The second edge does not have to be connected to the first edge.

Step 3: Choose another edge of minimum weight from the remaining edges, but do not select any edge that creates a cycle in the subgraph you are creating.

Step 4: Repeat step 3 until all the vertices of the original graph are included and you have a spanning tree.

Example 12.52

Using Kruskal’s Algorithm

A computer network will be set up with six devices. The vertices in the graph in Figure 12.229 represent the devices, and the edges represent the cost of a connection. Find the network configuration that will cost the least. What is the total cost?

Figure 12.229 Graph of Network Connection Costs

Solution

A minimum spanning tree will correspond to the network configuration of least cost. We will use Kruskal’s algorithm to find one. Since the graph has six vertices, the spanning tree will have six vertices and five edges.

Step 1: Choose an edge of least weight. We have sorted the weights into numerical order. The least is $100. The only edge of this weight is edge AF as shown in Figure 12.230.

Figure 12.230 Step 1 Select Edge AF

Step 2: Choose the edge of least weight of the remaining edges, which is BD with $120. Notice that the two selected edges do not need to be adjacent to each other as shown in Figure 12.231.

Figure 12.231 Step 2 Select Edge BD

Step 3: Select the lowest weight edge of the remaining edges, as long as it does not result in a cycle. We select DF with $150 since it does not form a cycle as shown in Figure 12.232.

Figure 12.232 Step 3 Select Edge DF

Repeat Step 3: Select the lowest weight edge of the remaining edges, which is BE with $160 and it does not form a cycle as shown in Figure 12.233. This gives us four edges so we only need to repeat step 3 once more to get the fifth edge.

Figure 12.233 Repeat Step 3 Select Edge DF

Repeat Step 3: The lowest weight of the remaining edges is $170. Both BF and CE have a weight of $170, but BF would create cycle (b, d, f) and there cannot be a cycle in a spanning tree as shown in Figure 12.234.

Figure 12.234 Repeat Step 3 Do Not Select Edge BF

So, we will select CE, which will complete the spanning tree as shown in Figure 12.235.

Figure 12.235 Repeat Step 3 Select Edge CE

The minimum spanning tree is shown in Figure 12.236. This is the configuration of the network of least cost. The spanning tree has a total weight of $\$100+\$120+\$150+\$160+\$170=\$700$, which is the total cost of this network configuration.

Figure 12.236 Final Minimum Spanning Tree

Your Turn 12.52

1.

Find a minimum spanning tree for the weighted graph. Give its total weight.

Weighted Graph