Skip to main content

Section 2.1 Eulerian Circuits

Problem 2.1.1.

The edges of the graph in Figure 2.1.2 represent bridges connecting plots of land represented by the vertices. Try to find a way to walk across all the bridges using each bridge exactly once starting and ending at the same location. Vertices may be re-used.

Solution.

There are really only two possibilities to walk along these paths.

  1. We can choose to first walk along the lines where each vertex is adjacent to two other vertices. In other words, we start at any points but \(v_1\) and \(v_4\text{.}\) If we do this, however; we miss either one or two lines to traverse, before completing a cycle.

  2. We can choose to start at either \(v_1\) or \(v_4\text{,}\) each adjacent to three vertices, including each other. Again we miss lines to traverse. Either we miss the line connecting these two vertices, or if we start at the line connecting them, we miss the lines on either side of the graph. If we do decide to cross out initial vertex a second time to traverse the missing lines, we are forced to cross the line connecting the \(v_1\) and \(v_4\) a second time, which we are prohibited from doing.

In either of these two cases, we would be forced to reuse edges (lines or bridges) to return to our original location.

Figure 2.1.2. Bridges of Königsberg

Problem 2.1.3.

Repeat this exercise, but this time try to find a way to walk across all the bridges using each bridge exactly once starting and ending anywhere. Vertices may be re-used.

Solution.

Definition 2.1.4. Eulerian circuit.

A sequence of edges in a graph \(G\) is an Eulerian circuit if and only if they are of the form

\(v_1v_2, v_2v_3, v_3v_4, \ldots v_{k-1}v_k, v_kv_1\) such that the graph \(G\) has \(k\) edges and each edge is unique in the sequence.

Definition 2.1.5. Eulerian walk.

A sequence of edges in a graph \(G\) is an Eulerian walk if and only if they are of the form

\(v_1v_2, v_2v_3, v_3v_4, \ldots v_{k-1}v_k\) such that the graph \(G\) has \(k-1\) edges and each edge in the sequence is unique.

Definition 2.1.6. Degree.

The degree of a vertex is the number of edges incident with that vertex.

Problem 2.1.7.

For each of the graphs in Figure 2.1.8 list the degrees of each vertex. Then list the degrees of each vertex in the graph in descending order.

Solution.

  • A: 3, B: 2, C: 3, D: 2

  • A: 3, C: 3, B: 2, D: 2

  • A: 2, B: 2, C: 2, D: 2, E: 4

  • E: 4, A: 2, B: 2, C: 2, D: 2

  • Mon: 0, Tues: 4, Wed: 1,Thurs: 3, Fri: 0, Sat: 3, Sun: 3

  • Tues: 4, Thurs: 3, Sat: 3, Sun: 3, Wed: 1, Mon: 0, Fri: 0

Graph 1

Graph 2

Graph 3

\(\left( \begin{array}{*{5}{c}} 0 \amp 1 \amp 0 \amp 0 \amp 1 \\ 1 \amp 0 \amp 0 \amp 0 \amp 1 \\ 0 \amp 0 \amp 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 1 \amp 0 \amp 1 \\ 1 \amp 1 \amp 1 \amp 1 \amp 0 \end{array} \right)\)

The days of the week are adjacent if and only if they share a letter.

Figure 2.1.8. Degree Sequences

Problem 2.1.9.

For each of the following sequences, if possible build a graph such that each number represents the degree of a distinct vertex.

(f)

5,4,3,2,1

Solution.

Not possible, as there is either a vertex with extra edges or an edge with no ending vertex.

(g)

1,1,1

Solution.

Not possible, as only 1 valid edge can be made.

Proof.

The amount of edges in a graph is half the sum of the degrees of all vertices in that graph. For Graph G it has 5 vertices and 8 edges. The degrees of the vertices are: 4,4,3,3,2. The sum of those numbers is 16. Half of 16 is 8.

This is because for every one vertex added to a graph, the sum of all degrees increases by 2. Since adding a vertex adds 1 degree to that vertex, but also to an incident vertex. So for every edge, there are two vertices sharing that edge. Adding one edge, adds 2 to the sum of all of the degrees of the vertices.

Proof.

We will prove by contradiction that the number of vertices of odd degree is even. As demonstrated by Lemma 2.1.10, the sum of the degrees of vertices of a graph will be twice the number of edges. The sum of degree of vertices will be an even number. If there are an odd number of vertices with an odd number of degrees, the sum of degrees will be odd, which is not allowed.

Proof.

We prove the contrapositive. Let \(G\) be a non-trivial connected graph that does not contain a cycle. (If \(G\) is not connected, apply the argument to a connected subgraph of \(G\text{.}\)) We show that \(G\) contains a vertex of degree 1.

Since \(G\) is connected and contains no cycles, for every pair of nonadjacent vertices \(u,v\) in \(G\text{,}\) there is a \(uv\)-path of length greater than 1. Moreover, the converse is also true: if there is a \(uv\)-path of length greater than 1, then \(u\) and \(v\) are nonadjacent (for otherwise there would be a cycle.)

Consider the longest path in \(G\text{.}\) If this path is of length 1, we are done, since this is only possible if there are vertices of degree 1. So we can assume the path is of length 2 or greater. It follows that the endpoints of this path are not adjacent. Moreoever, since it is a longest path, there are no further vertices adjacent to either of the endpoints. Therefore, the endpoints are of degree 1.

Proof.

We will prove that a connected graph with exactly two vertices of odd degree has an Eulerian walk. Recall that an Eulerian walk traverses every edge once, and vertices can be crossed multiple time. Vertices of even degree have an "entrance" and an "exit" - meaning, we can cross the vertex (possible multiple times) if it has an even degree. The Eulerian walk must begin and end at the odd-degreed vertices. If we try to contradict this, and begin our walk on an even-degreed vertex, we must backtrack to traverse the edge connected to a vertex with odd degree, which is not allowed.

Proof.

Suppose a connected graph has exactly two vertices of odd degree. There are two ways to get this kind of construction:

First, one can start with a graph E where all the vertices are of even degree (which contains an Eulerian circuit), then choose two vertices that are not adjacent (call them \(v_s\) and \(v_e\)) and connect them with an edge (call it \(q_n\)). Call this new graph G. By following the Eulerian circuit that existed in E, one can traverse from \(v_s\) to \(v_s\) and exhaust every edge except \(q_n\text{.}\) However, traversing \(q_n\) exhausts every edge and terminates at \(v_e\text{,}\) an Eulerian walk.

An example of using the first construction. Clearly the first image contains an Eulerian Circuit that begins and ends at A. By adding the green edge, an Eulerian walk can be found that follows the circuit from the first image, then simply follows the edge connecting \(A\) and \(D\)

Alternatively, such a construction can be made by starting with a graph E where all of the vertices are of even degree and E has at least one edge (clearly, E has an Eulerian circuit). Form a graph G by deleting one edge along the circuit. By following the remains of E's circuit as far as possible, one can see that G contains an Eulerian walk.

An example of using the second construction. The first image depicts an Eulerian circuit that exists in a graph with all vertices of even degree. By deleting the edge that joins \(A\) and \(F\text{,}\) we get an Eulerian walk that starts at \(A\) and ends at \(F\text{.}\)

Either way, both constructions result in a graph that possesses an Eulerian walk. Therefore, any connected graph with exactly two vertices of odd degree has an Eulerian Walk.