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 12.231 is a tree.
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 12.232. If a graph is not a tree, explain how you know.
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
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 12.234 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 12.235.
Example 12.47
Identifying Types of Trees
Each graph in Figure 12.236 is one of the special types of trees we have been discussing. Identify the type of tree.
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 12.237. 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).
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 12.238 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 12.238.
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 12.239.
FORMULA
The number of edges in a tree graph with $n$ vertices is $n1$.
A connected graph with n vertices and $n1$ edges is a tree graph.
Example 12.48
Exploring Characteristics of Trees
Use Graphs I and J in Figure 12.240 to answer each question.
 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?
Solution
 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.
Your Turn 12.48
Who Knew?
Graph Theory in the Movies
In the 1997 film Good Will Hunting, the main character, Will, played by Matt Damon, solves what is supposed to be an exceptionally difficult graph theory problem, “Draw all the homeomorphically irreducible trees of size $n=10$.” That sounds terrifying! But don’t panic. Watch this great Numberphile video to see why this is actually a problem you can do at home!
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 12.241, in which each device is connected directly to every other device in the network.
The mesh topology for four devices could be represented by the complete Graph A_{1} in Figure 12.242 where the vertices represent the devices, and the edges represent network connections. However, the devices could be networked using fewer connections. Graphs A_{2}, A_{3,} and A_{4} of Figure 12.242 show configurations in which three of the six edges have been removed. Each of the Graphs A_{2}, A_{3} and A_{4} in Figure 12.242 is a tree because it is connected and contains no cycles. Since Graphs A_{2}, A_{3} and A_{4} are also subgraphs of Graph A_{1} that include every vertex of the original graph, they are also known as spanning trees.
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 12.243 to determine which of graphs M_{1}, M_{2}, M_{3}, and M_{4}, are spanning trees of Q.
Solution
 Graph M_{1} is not a spanning tree of Graph Q because it has a cycle (c, d, f, e).
 Graph M_{2} is a spanning tree of Graph Q because it has all the original vertices, no vertices are adjacent in M_{2} that weren’t adjacent in Graph Q, Graph M_{2} is connected and it contains no cycles.
 Graph M_{3} is not a spanning tree of Graph Q because vertices a and f are adjacent in Graph M_{3} but not in Graph Q.
 Graph M_{4} is not a spanning tree of Graph Q because it is not connected.
So, only graph M_{2} is a spanning tree of Graph Q.
Your Turn 12.49

True

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 12.245.
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 12.246.
Repeat this process until all vertices have been visited as shown in Figure 12.247.
The end result is a tree that spans the entire graph as shown in Figure 12.248.
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 12.249.
Solution
Your Turn 12.50
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 12.253.
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 12.254.
Example 12.51
Removing Edges to Find Spanning Trees
Use the graph in Figure 12.255 to answer each question.
 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.
Solution
 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 12.256.
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 12.257.
Your Turn 12.51
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 fourcycles.
“…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. 4491, 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 12.259 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?
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 12.260.
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 12.261.
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 12.262.
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 12.263. This gives us four edges so we only need to repeat step 3 once more to get the fifth edge.
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 12.264.
So, we will select CE, which will complete the spanning tree as shown in Figure 12.265.
The minimum spanning tree is shown in Figure 12.266. 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.
Your Turn 12.52
Check Your Understanding

True

False

True

False
Section 12.10 Exercises
A  B  C  D  

A    125  320  275 
B  125    110  540 
C  320  110    1,010 
D  275  540  1,010   
V  W  X  Y  Z  

V    2  4  9  10 
W  2    6  8  No Path 
X  4  6    7  No Path 
Y  9  8  7    5 
Z  10  No Path  No Path  5   