### Learning Objectives

After completing this section, you should be able to:

- Describe and identify Euler trails.
- Solve applications using Euler trails theorem.
- Identify bridges in a graph.
- Apply Fleuryâ€™s algorithm.
- Evaluate Euler trails in real-world applications.

We used Euler circuits to help us solve problems in which we needed a route that started and ended at the same place. In many applications, it is not necessary for the route to end where it began. In other words, we may not be looking at circuits, but trails, like the old Pony Express trail that led from Sacramento, California in the west to St. Joseph, Missouri in the east, never backtracking.

### Euler Trails

If we need a trail that visits every edge in a graph, this would be called an Euler trail. Since trails are walks that do not repeat edges, an Euler trail visits every edge exactly once.

### Example 12.29

#### Recognizing Euler Trails

Use Figure 12.150 to determine if each series of vertices represents a trail, an Euler trail, both, or neither. Explain your reasoning.

*a*â†’*b*â†’*e*â†’*g*â†’*f*â†’*c*â†’*d*â†’*e**a*â†’*b*â†’*e*â†’*g*â†’*f*â†’*c*â†’*d*â†’*e*â†’*b*â†’*a*â†’*d*â†’*g**g*â†’*d*â†’*a*â†’*b*â†’*e*â†’*d*â†’*c*â†’*f*â†’*g*â†’*e*

#### Solution

- It is a trail only. It is a trail because it is a walk that doesnâ€™t cover any edges twice, but it is not an Euler trail because it didnâ€™t cover edges
*ad*or*dg*. - It is neither. It is not a trail because it visits
*ab*and*be*twice. Since it is not a trail, it cannot be an Euler trail. - It is both. It is a trail because it is a walk that doesnâ€™t cover any edges twice, and it is an Euler trail because it visits all the edges.

### Your Turn 12.29

### The Five Rooms Puzzle

Just as Euler determined that only graphs with vertices of even degree have Euler circuits, he also realized that the only vertices of odd degree in a graph with an Euler trail are the starting and ending vertices. For example, in Figure 12.150, Graph *H* has exactly two vertices of odd degree, vertex *g* and vertex *e*. Notice the Euler trail we saw in Excercise 3 of Example 12.29 began at vertex *g* and ended at vertex *e*.

This is consistent with what we learned about vertices off odd degree when we were studying Euler circuits. We saw that a vertex of odd degree couldn't exist in an Euler circuit as depicted in Figure 12.152. If it was a starting vertex, at some point we would leave the vertex and not be able to return without repeating an edge. If it was not a starting vertex, at some point we would return and not be able to leave without repeating an edge. Since the starting and ending vertices in an Euler trail are not the same, the start is a vertex we want to leave without returning, and the end is a vertex we want to return to and never leave. Those two vertices must have odd degree, but the others cannot.

Letâ€™s use the Euler trail theorem to solve a puzzle so you can amaze your friends! This puzzle is called the â€śFive Rooms Puzzle.â€ť Suppose that you were in a house with five rooms and the exterior. There is a doorway in every shared wall between any two rooms and between any room and the exterior as shown in Figure 12.153. Could you find a route through the house that passes through each doorway exactly once?

Letâ€™s represent the puzzle with a graph in which vertices are rooms (or the exterior) and an edge indicates a door between two rooms as shown in Figure 12.154.

To pass through each doorway exactly once means that we cross every edge in the graph exactly once. Since we have not been asked to start and end at the same position, but to visit each edge exactly once, we are looking for an Euler trail. Letâ€™s check the degrees of the vertices.

Since there are more than two vertices of odd degree as shown in Figure 12.155, the graph of the five rooms puzzle contains no Euler path. Now you can amaze and astonish your friends!

### Bridges and Local Bridges

Now that we know which graphs have Euler trails, letâ€™s work on a method to find them. The method we will use involves identifying bridges in our graphs. A bridge is an edge which, if removed, increases the number of components in a graph. Bridges are often referred to as cut-edges. In Figure 12.156, there are several examples of bridges. Notice that an edge that is not part of a cycle is always a bridge, and an edge that is part of a cycle is never a bridge.

Edges *bf*, *cg*, and *dg* are â€śbridgesâ€ť

The graph in Figure 12.156 is connected, which means it has exactly one component. Each time we remove one of the bridges from the graph the number of components increases by one as shown in Figure 12.157. If we remove all three, the resulting graph in Figure 12.157 has four components.

In sociology, bridges are a key part of social network analysis. Sociologists study two kinds of bridges: local bridges and regular bridges. Regular bridges are defined the same in sociology as in graph theory, but they are unusual when studying a large social network because it is very unlikely a group of individuals in a large social network has only one link to the rest of the network. On the other hand, a local bridge occurs much more frequently. A local bridge is a friendship between two individuals who have no other friends in common. If they lose touch, there is no single individual who can pass information between them. In graph theory, a local bridge is an edge between two vertices, which, when removed, increases the length of the shortest path between its vertices to more than two edges. In Figure 12.158, a local bridge between vertices *b* and *e* has been removed. As a result, the shortest path between *b* and *e* is *b* â†’ *i* â†’ *j* â†’ *k* â†’ *e*, which is four edges. On the other hand, if edge *ab* were removed, then there are still paths between *a* and *b* that cover only two edges, like *a* â†’ *i* â†’ *b*.

The significance of a local bridge in sociology is that it is the shortest communication route between two groups of people. If the local bridge is removed, the flow of information from one group to another becomes more difficult. Letâ€™s say that vertex *b* is Brielle and vertex *e* is Ella. Now, Brielle is less likely to hear about things like job opportunities, that Ella many know about. This is likely to impact Brielle as well as the friends of Brielle.

### Example 12.30

#### Identifying Bridges and Local Bridges

Use the graph of a social network in Figure 12.159 to answer each question.

- Identify any bridges.
- If all bridges were removed, how many components would there be in the resulting graph?
- Identify one local bridge.
- For the local bridge you identified in part 3, identify the shortest path between the vertices of the local bridge if the local bridge were removed.

#### Solution

- The edges
*ku, gh*, and*hi*are bridges. - If the bridges were all removed, there would be four components in the resulting graph, {
*i*}, {*h*}, {*u*,*v*,*w*,*x*}, and {*a*,*b*,*c*,*d*,*e*,*f*,*g*,*j*,*k*,*m*,*n*,*o*,*p*,*q*,*r*,*s*,*t*} as shown in Figure 12.160. - Three local bridges are
*dn, ef*, and*qt*, among others. - If
*dn*were removed, the shortest path between*d*and*n*would be*d*â†’*e*â†’*f*â†’*j*â†’*o*â†’*m*â†’*n*.

### Your Turn 12.30

### Finding an Euler Trail with Fleuryâ€™s Algorithm

Now that we are familiar with bridges, we can use a technique called Fleuryâ€™s algorithm, which is a series of steps, or algorithm, used to find an Euler trail in any graph that has exactly two vertices of odd degree. Here are the steps involved in applying Fleuryâ€™s algorithm.

Here are the steps involved in applying Fleuryâ€™s algorithm.

**Step 1:** Begin at either of the two vertices of odd degree.

**Step 2:** Remove an edge between the vertex and any adjacent vertex that is NOT a bridge, unless there is no other choice, making a note of the edge you removed. Repeat this step until all edges are removed.

**Step 3:** Write out the Euler trail using the sequence of vertices and edges that you found. For example, if you removed *ab, bc, cd, de*, and *ef*, in that order, then the Euler trail is *a* â†’ *b* â†’ *c* â†’ *d* â†’ *e* â†’ *f*.

Figure 12.161 shows the steps to find an Euler trail in a graph using Fleuryâ€™s algorithm.

The Euler trail that was found in Figure 12.161 is *t* â†’ *v* â†’ *w* â†’ *u* â†’ *t* â†’ *w* â†’ *y* â†’ *x* â†’ *v*.

### Example 12.31

#### Finding an Euler Trail with Fleuryâ€™s Algorithm

Use Fleuryâ€™s Algorithm to find an Euler trail for Graph *J* in Figure 12.162.

#### Solution

**Step 1:** Choose one of the two vertices of odd degree, *c* or *f*, as your starting vertex. We will choose *c*.

**Step 2:** Remove edge *ca, cb*, or *cd*. None of these are cut edges so we can select any of the three. We will choose *cb* as shown in Figure 12.163 to be the first edge removed.

**Repeat Step 2** The next choice is to remove edge *ba, bd*, or *bf* as shown in Figure 12.163, but *bf* is not an option since it is a bridge. We will choose *ba* as shown in Figure 12.164 to be the second edge removed.

**Repeat Step 2** for the third, fourth, fifth, sixth, and seventh edges. As shown in Figure 12.164, until we get to the seventh edge there is only one option each time, *ac, cd, db*, and *bf* in that order. For the seventh edge, we must choose between *fe* and *fg*. Neither of these are bridges. We choose *fe*. Figure 12.165 shows that *ac, cd, db, bf,* and *fe* have been removed.

**Repeat Step 2** for the eight, ninth, tenth, and eleventh edges. As shown in Figure 12.165, there is only one option for each of these edges, *eh, hi, ig*, and *gf*, in that order.

**Step 3:** Write out the Euler trail using the vertices in the sequence that the edges were removed. We removed *cb, ba, ac, cd, db, bf, fe, eh, hi, ig*, and *gf*, in that order. The Euler trail is *c* â†’ *b* â†’ *a* â†’ *c* â†’ *d* â†’ *b* â†’ *f* â†’ *e* â†’ *h* â†’ *i* â†’ *g* â†’ *f*.

### Checkpoint

*TIP! To avoid errors, count the number of edges in your graph and make sure thatyour Euler trail represents that number of edges.*

### Your Turn 12.31

*L*to fill in the blanks to complete the steps in Fleuryâ€™s algorithm.

*sq*is the first edge removed, the three options for the second edge to be removed are

*qr*, ___, and ___; however, ___ cannot be chosen because it is a ________________.

*qr*is the second edge removed, the next four edges to be removed must be ___, ___, ___, and ___, in that order.

*qn*is removed, the three options for the next edge to be removed are

*no*, ___, and ___, however, ___ cannot be chosen because it is a _____________.

*no*is the next edge removed, the last four edges removed will be ___, ___, ___, and ___, in that order.

In the previous section, we found Euler circuits using an algorithm that involved joining circuits together into one large circuit. You can also use Fleuryâ€™s algorithm to find Euler circuits in any graph with vertices of all even degree. In that case, you can start at any vertex that you would like to use.

**Step 1:** Begin at any vertex.

**Step 2:** Remove an edge between the vertex and any adjacent vertex that is NOT a bridge, unless there is no other choice, making a note of the edge you removed. Repeat this step until all edges are removed.

**Step 3:** Write out the Euler circuit using the sequence of vertices and edges that you found. For example, if you removed *ab, bc, cd, de*, and *ea*, in that order, then the Euler circuit is *a* â†’ *b* â†’ *c* â†’ *d* â†’ *e* â†’ *a*.

### Checkpoint

*IMPORTANT! Since a circuit is a closed trail, every Euler circuit is also an Euler trail, but when we say Euler trail in this chapter, we are referring to an open Euler trail that begins and ends at different vertices.*

### Example 12.32

#### Finding an Euler Circuit or Euler Trail Using Fleuryâ€™s Algorithm

Use Fleuryâ€™s algorithm to find either an Euler circuit or Euler trail in Graph *G* in Figure 12.167.

#### Solution

Graph *G* has all vertices of even degree so it has an Euler circuit.

**Step 1:** Choose any vertex. We will choose vertex *j*.

**Step 2:** Remove one of the four edges that meet at vertex *j*. Since *jn* is a bridge, we must remove either *jh, ji*, or *jk.* We remove *ji* as shown in Figure 12.168.

**Repeat Step 2:** Since *id* is a bridge, we can remove either *ih* or *ik* next. We remove *ih*, and then the only option is to remove *hj* as shown in Figure 12.169.

**Repeat Step 2:** Since *jn* is a bridge, the next edge removed must be *jk*, and then the only option is to remove *ki* followed by *id* as shown in Figure 12.169. Even though *id* is a bridge, it can be removed because it is the only option at this point. Figure 12.170 shows Graph *G* with these additional edges removed.

**Repeat Step 2:** Choose any one of the edges *db, dc*, or *de*. We remove *dc* as shown in Figure 12.171.

**Repeat Step 2:** Since *co* is a bridge, choose *cb* next. We remove *cb*, then *bd*, and then *de* as shown in Figure 12.172.

**Repeat Step 2:** Next, remove *ec* and *co*. Then choose any of *op, pn*, or *om*. We remove on as shown in Figure 12.173.

**Repeat Step 2:** Next, remove either *nm*, *np*, or *nj*, but *nj* is a So, we remove *nm* as shown in Figure 12.174.

**Repeat Step 2:** Next, remove *mo, op, pn*, and *nj*. And we are done!

**Step 3:** Notice that the algorithm brought us back to the vertex where we started, forming an Euler circuit. Write out the Euler circuit:

*j* â†’ *i* â†’ *h* â†’ *j* â†’ *k* â†’ *i* â†’ *d* â†’ *c* â†’ *b* â†’ *d* â†’ *e* â†’ *c* â†’ *o* â†’ *n* â†’ *m* â†’ *o* â†’ *p* â†’ *n* â†’ *j*

### Your Turn 12.32

### WORK IT OUT

We have discussed a lot of subtle concepts in this section. Letâ€™s make sure we are all on the same page. Work with a partner to explain why each of the following facts about bridges are true. Support your explanations with definitions and graphs.

- When a bridge is removed from a graph, the number of components increases.
- A bridge is never part of a circuit.
- An edge that is part of a triangle is never a local bridge.

### Check Your Understanding

### Section 12.6 Exercises

*ab*is a bridge

*ef*is a bridge

*ab*is a local bridge

*ef*is a local bridge

*A*,

*w*â†’

*x*â†’

*y*â†’

*z*â†’

*w*â†’

*u*â†’

*t*â†’

*s*â†’

*v*â†’

*u*

*A*,

*u*â†’

*v*â†’

*s*â†’

*t*â†’

*u*â†’

*w*â†’

*z*â†’

*y*â†’

*x*â†’

*w*

*A*,

*s*â†’

*t*â†’

*u*â†’

*v*â†’

*u*â†’

*w*â†’

*z*â†’

*y*â†’

*x*â†’

*w*

*A*,

*w*â†’

*x*â†’

*y*â†’

*z*â†’

*w*â†’

*v*â†’

*u*â†’

*t*â†’

*s*â†’

*v*

*B*,

*u*â†’

*v*â†’

*w*â†’

*x*â†’

*r*â†’

*u*â†’

*t*â†’

*s*â†’

*y*â†’

*z*â†’

*u*

*B*,

*v*â†’

*w*â†’

*x*â†’

*r*â†’

*u*â†’

*z*â†’

*y*â†’

*s*â†’

*t*â†’

*u*

*C*,

*s*â†’

*t*â†’

*u*â†’

*v*â†’

*w*â†’

*x*â†’

*s*

*C*,

*t*â†’

*u*â†’

*x*â†’

*w*â†’

*u*â†’

*s*â†’

*t*â†’

*v*â†’

*w*

*D*,

*t*â†’

*r*â†’

*s*â†’

*t*â†’

*u*â†’

*v*â†’

*t*â†’

*x*â†’

*v*â†’

*w*â†’

*x*â†’

*y*â†’

*z*â†’

*x*

*D*,

*x*â†’

*v*â†’

*w*â†’

*x*â†’

*y*â†’

*z*â†’

*x*â†’

*t*â†’

*r*â†’

*s*â†’

*t*â†’

*u*â†’

*v*â†’

*t*

*A*

*B*

*C*

*D*

*A*

*B*

*C*

*D*

*Q*

*R*

*S*

A graph in which each vertex represents a space on a five-by-six game board and each edge represents a move a knight could make is shown in the figure.

A knightâ€™s tour is a sequence of moves by a knight on a chessboard (of any size) such that the knight visits every square exactly once. If the knightâ€™s tour brings the knight back to its starting position on the board, it is called a closed knightâ€™s tour. Otherwise, it is called an open knightâ€™s tour. Determine if the closed knightâ€™s tour in the figure is most accurately described as a trail, a circuit, an Euler trail, or an Euler circuit of the graph of all possible knight moves. Explain your reasoning.