In the graph shown in Fig 10.1, ABCEFCBD is a walk of length 7 which is in Java Maker QR Code JIS X 0510 in Java In the graph shown in Fig 10.1, ABCEFCBD is a walk of length 7 which is

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EXAMPLE 1 In the graph shown in Fig 10.1, ABCEFCBD is a walk of length 7 which is use spring framework qr bidimensional barcode drawer toaccess qr-code with java QR Code Safty neither a trail nor a path; A qr bidimensional barcode for Java BCEFCD is a trail, but not a path; ABCEFCDBA is a closed walk which is not a circuit; BCEFCDB is a circuit which is not a cycle; and BCDB is a 3-cycle and hence an odd cycle. The closed walk CEFCBDC is not a cycle because the first and last vertex appears a third time. A.

10 . Paths and Circuits Verify each of the assertions just made. While, strictly speaking, a walk should be specified by an alternating sequence of vertices and edges, it is often enough to specify only the vertices, as long as consecutive vertices are adjacent, as we have just done, or only the edges, as long as consecutive edges are adjacent." An important type of circuit, known as an Eulerian circuit, is one which passes through every edge of a pseudograph.

. An Eulerian circuit in a pseu jsp Denso QR Bar Code dograph is a circuit which contains every vertex and every edge. An Eulerian pseudograph is a pseudograph which contains an Eulerian circuit. Note the reference to vertices in the definition of Eulerian circuit, which is crucial to the theory.

See Exercise 5.. EXAMPLE 2 In the graph of Fig 10.1, the QR Code JIS X 0510 for Java circuit ABCEFCDA is not Eulerian because it does not contain the edge BD. As we shall soon see, this graph possesses no Eulerian circuit; it is not an Eulerian graph.

A Figure 10.2 further illustrates the difference between a circuit and an Eulerian circuit. The circuit ABCDEFGHFA is not Eulerian since, while it encompasses all vertices, it omits four edges.

The graph is Eulerian, however: ABCDEFGHFADBEA is an Eulerian circuit. A. H A B C G E D EXAMPLE 3 An Eulerian graph. Eulerian circuits are named, swing QR Code ISO/IEC18004 of course, after LUonhard Euler, the solver of the Konigsberg Bridge Problem, and their study is motivated by that problem. In Section 9.1, we saw that to follow the desired route over the bridges of Konigsberg, you have to choose a vertex in the pseudograph of Fig 9.

2 and find a walk which includes all the edges exactly once and leads back to the chosen vertex. With our present terminology, the Konigsberg Bridge Problem asks if the pseudograph is Eulerian..

In a graph, there is never a problem specifying just vertices or just edges. 10.1 Eulerian Circuits A B C D E A graph which is not Eulerian. In Theorem 10.1.4, we present the remarkably simple test which Euler found for the presence of an Eulerian circuit.

First, since an Eulerian circuit provides a walk between every pair of vertices, any Eulerian pseudograph must be connected, in the following sense.. MPOM A pseudograph is connected if and only if there exists a walk between any two vertices. As we ask you to show in Exercise 16, if a graph is connected, then there is actually a path between any two vertices, not just a walk. An Eulerian graph must be more than just connected, however, as the graph in Fig 10.

3 illustrates. The basic difficulty with this graph is that there is only one edge incident with A. Any circuit that begins at A cannot return to A without using this edge again, and any circuit which begins at a vertex other than A and attempts to include all edges, after using the edge BA, has to repeat it en route back to the starting vertex.

The degrees of the vertices play a role in determining whether or not a graph is Eulerian. It is not hard to see that the degrees of the vertices of an Eulerian graph must be even. In essence, we have already given the argument.

In walking along an Eulerian circuit, every time we meet a vertex (other than the one where we started), either we leave on a loop and return immediately, never traversing that loop again, or we leave on an edge different from that by which we entered and traverse neither edge again. So the edges (other than loops) incident with any vertex in the middle of the circuit can be paired. So also can the edges incident with the first (and last) vertex since the edge by which we left it at the beginning can be paired with the edge by which we returned at the end.

Thus, an Eulerian graph must not only be connected, but also have vertices of even degree. Conversely, a connected graph all of whose vertices are even must be Eulerian. To see why, it will be helpful to examine again the graph of Fig 10.

2 (which is connected and has only even vertices) and to try to construct an Eulerian circuit with a strategy that might apply more generally. We attempt to find an Eulerian circuit starting at A. To begin, we find some circuit which starts and ends at A, for instance, the circuit Cl: ABCDEFA.

This circuit obviously is not Eulerian because it misses lots of edges. If we delete the edges of C1 from the graph as well as vertex C, which is isolated after the edges of Cl have been removed, we are left with the graph g, on the left of Fig 10.4.

This graph gi has a circuit C, FGH, which is connected to C1 at vertex F. Thus, it can be used to enlarge Cl as follows: Start at A, follow C1 as far.
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