12.10 Trees

A row of trees with no leaves. — **Figure 12.203** In graph theory, graphs known as trees have structures in common with live trees.In graph theory, graphs known as trees have structures in common with live trees. (credit: “Row of trees in Roslev” by AKA CJ/Flickr, Public Domain)

Learning Objectives

After completing this section, you should be able to:

Describe and identify trees.
Determine a spanning tree for a connected graph.
Find the minimum spanning tree for a weighted graph.
Solve application problems involving trees.

We saved the best for last! In this last section, we will discuss arguably the most fun kinds of graphs, trees. Have you every researched your family tree? Family trees are a perfect example of the kind of trees we study in graph theory. One of the characteristics of a family tree graph is that it never loops back around, because no one is their own grandparent!

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.

Three graphs. Graph T has 15 vertices. The edges are as follows: a b, b c, c d, c i, i j, j k, k o, d e, d l, e n, e m, b f, f g, and g h. Graph P has 6 vertices. The edges are as follows: t s, s r, r q, q p, and p o. Graph S has 7 vertices. The edges are as follows: u t, u v, u w, u x, u y, and u z. — **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.

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.

Graph F has 8 vertices. The vertices are labeled a to h. Edges connect b a, b d, b c, f e, f h, and f g. — **Figure 12.206** Forest Graph F

Six graphs. Graph S has 7 vertices. The edges are a b, b f, f g, f c, b c, b d, and d e. Graph N has 6 vertices. The edges are I j, I h, l k, and l m. The edges, I j, and l k intersect each other. Graph P has 6 vertices. The edges are s t, s r, r q, q p, and p o. Graph Z has 12 vertices. The edges are a b, a c, a d, a e, a f, a g, a h, h i, c j, e k, and k l. Graph C has 14 vertices. The edges are m n, n z, n o, o r, o p, o x, o y, p y, p x, p s, p t, p u, p q, p v, and p w. Graph L has 26 vertices. The edges are f q, f t, f s, f e, e o, e p, p z, e d, d m, m v, d n, n y, d c, c k, k u, c l, l k, c b, b i, i t, b j, j w, b a, b g, and b h. — **Figure 12.207** Six Types of Trees

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).

Three graphs. Graph T has 15 vertices. The edges are as follows: a b, b c, c d, c i, i j, j k, k o, d e, d l, e n, e m, b f, f g, and g h. The edges, c i, and i j are in blue. A dashed edge connects b and j. Graph P has 6 vertices. The edges are as follows: t s, s r, r q, q p, and p o. The edges, t s, and s r are in blue. A dashed edge connects t and r. Graph S has 7 vertices. The edges are as follows: u t, u v, u w, u x, u y, and u z. The edges, u t, and u v are in blue. A dashed edge connects t and 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.

Three graphs. Graph T has 15 vertices. The edges are as follows: a b, b c, c d, c i, i j, j k, k o, d e, d l, e n, e m, b f, f g, and g h. The edges, d c, and b f are in dashed lines. The graph is separated into three blocks. Graph P has 6 vertices. The edges are as follows: t s, s r, r q, q p, and p o. The edge, q r is in dashed lines. The graph is separated into two blocks. Graph S has 7 vertices. The edges are as follows: u t, u v, u w, u x, u y, and u z. The edge, u w is in dashed lines. The graph is separated into two blocks. — **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.

Six graphs. Graph T has 15 vertices. The edges are as follows: a b, b c, c d, c i, i j, j k, k o, d e, d l, e n, e m, b f, f g, and g h. Graph P has 6 vertices. The edges are as follows: t s, s r, r q, q p, and p o. Graph S has 7 vertices. The edges are as follows: u t, u v, u w, u x, u y, and u z. Graph Z has 12 vertices. The edges are a b, a c, a d, a e, a f, a g, a h, h i, c j, e k, and k l. Graph C has 14 vertices. The edges are m n, n z, n o, o r, o p, o x, o y, p y, p x, p s, p t, p u, p q, p v, and p w. Graph L has 26 vertices. The edges are f q, f t, f s, f e, e o, e p, p z, e d, d m, m v, d n, n y, d c, c k, k u, c l, l k, c b, b i, i t, b j, j w, b a, b g, and b h. — **Figure 12.211** Number of Vertices and Edges in Trees vs. Other Graphs

Example 3 Exploring Characteristics of Trees

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

Figure 12.212 Graphs I and J

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

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.
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.
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.

Four illustrations represent the common network configurations. The first illustration represents mesh topology. Six computers are interconnected. The second illustration represents a ring topology. Five computers are connected in a ring. The third illustration represents star topology. A computer at the center is connected to five computers surrounding it. The fourth illustration represents tree topology. Two branches arise from a horizontal bus. Each branch has a computer at the center connected to five computers surrounding it. — **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.

Four graphs. Graph A 1 has four vertices: a, b, c, and d. The edges are a b, b d, d c, c a, a d, and b c. Graph A 2 has four vertices: a, b, c, and d. The edges are a b, b c, and c d. Graph A 3 has four vertices: a, b, c, and d. The edges are a b, a c, and c d. Graph A 4 has four vertices: a, b, c, and d. The edges are a b, a c, and a d. — **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.

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.

A graph with 23 vertices and 35 edges. Ten edges are highlighted in green. Two vertices are labeled started here and stopped here. — **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.

A graph with 23 vertices and 35 edges. Ten edges are highlighted in blue. Ten edges are highlighted in green. Two vertices are labeled started here and stopped here. — **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.

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

A graph with 23 vertices and 22 edges. — **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.

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.

Graph D has 10 vertices. The vertices are labeled from a to j. The edges are c d, c a, d a, a g, a h, g h, a b, b e, b f, e f, b i, b j, and I j. — **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.

Four graphs depict removing edges from graph D. In the first graph, the vertices are labeled from a to j. The edges are c d, c a, d a, a g, a h, g h, a b, b e, b f, e f, b i, b j, and i j. The edges, a c, e f, g h, and b j are shown in dashed lines. The second graph is the same as that of the first with edges, a c, e f, g h, and b j removed. In the third graph, the vertices are labeled from a to j. The edges are c d, c a, d a, a g, a h, g h, a b, b e, b f, e f, b i, b j, and i j. The edges, a c, a g, b f, and b i are shown in dashed lines. The fourth graph is the same as that of the first with edges, a c, a g, b f, and b i removed. — **Figure 12.224** Removing Four Edges from Graph D

Example 6 Removing Edges to Find Spanning Trees

Use the graph in Figure 12.225 to answer each question.

Figure 12.225 Graph V

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

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.
(a, c, d), (a, c, f), (a, d, c, f), and (b, e, h, i, g)
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 7 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

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

Key Terms

acyclic
tree
forest
path graph or linear graph
star tree
root
starlike tree
caterpillar tree
lobster tree
spanning tree
minimum spanning tree

Key Concepts

A brute force algorithm always finds the ideal solution but can be impractical whereas a greedy algorithm is efficient but usually does not lead to the ideal solution.
A Hamilton cycle of lowest weight is a solution to the traveling salesperson problem.
The brute force method finds a Hamilton cycle of lowest weight in a complete graph.
The nearest neighbor method is a greedy algorithm that finds a Hamilton cycle of relatively low weight in a complete graph.

Videos

Formulas

The number of edges in a tree graph with $n$ vertices is $n minus 1$ . A connected graph with n vertices and $n minus 1$ edges is a tree graph.

Projects

Everyone Gets a Turn! – Graph Colorings

Let’s put your knowledge of graph colorings to work! Your task is to plan a field day following these steps.

Select between seven and ten activities for your field day. You can look online for ideas.
Create a survey asking for the participants to select the three to five events in which they would most like to participate. Survey between seven and ten people.
Use the results of your survey to create a graph in which each vertex represents one of the events. A pair of vertices will be adjacent if there is at least one participant who would like to participate in both events.
Find a minimum coloring for the graph. Explain how you found it and how you know the chromatic number of the graph.
Use your solution to part d to determine the minimum number of timeslots you must use to ensure that everyone has the opportunity to participate in their top three events.
Find the complement of the graph you created. Explain what the edges in this graph represent.

A Beautiful Day in the Neighborhood – Euler Circuits

Let’s apply what you have learned to the community in which you live. Using resources such as your county’s property appraiser’s website, create a detailed graph of your neighborhood in which vertices represent turns and intersections. Represent a large enough part of your community to include no fewer than 10 intersections or turns. Then use your graph to answer the following questions.

Label the edges of your graph.
Determine if your graph is Eulerian. Explain how you know. If it is not, eulerize it.
Find an Euler circuit for your graph. Give the sequence of vertices that you found.
What does the Euler circuit you found in part c represent for your community?
Describe an application for which this Euler circuit might be used.

Dream Vacation – Hamilton Cycles and Paths

Where in the world would you like to travel most: the Eiffel Tower in Paris, a Broadway musical in New York city, a bike tour of Amsterdam, the Tenerife whale and dolphin cruises in the Canary Islands, the Giza Pyramid in Cairo, or maybe the Jokhang Temple in Tibet? Let's plan your dream vacation!

Which four destinations are at the top of your bucket list?
Draw a complete weighted graph with five vertices representing the four destinations and your home city, and the weights representing the cost of travel between cities.
Use a website (such as Travelocity) to find the best airfare between each pair of cities. List the airlines and flight numbers along with the prices. Include cost for ground transportation from the nearest airport if there is no airport at the destination you want to visit.
Use the nearest neighbor algorithm to find a Hamilton cycle of low weight beginning and ending in your hometown. What is the weight of this circuit and what does it represent?
Use the brute force method to find a Hamilton cycle of lowest weight beginning and ending in your hometown. What is the weight of this circuit? Is it the same or different from the weight of the Hamilton cycle you found in Exercise 4?
Suppose that instead of returning home, you planned to move to your favorite location on the list, but you wanted to stop at the other three destinations once along the way. Where would you move? List all Hamilton paths between your hometown and your favorite location.
Find the weights of all the Hamilton paths you found in Exercise 6.

Adapted from Contemporary Mathematics by OpenStax (openstax.org), licensed under CC BY-NC-SA 4.0. Changes were made. License: CC-BY-NC-SA-4.0.