Euler Circuits Question?
To traverse the circuit means to trace the circuit completely without lifting your pencil or retracing a line. You can go through the same intersection more than once though. To solve the circuit, count the number of lines connecting at each intersection. If all the intersections have an even number of lines coming into them, then you can start and finish the circuit at any point. If one of the intersections has an odd number of lines, then you must either start or finish at that intersection. If two of the intersections have odd number of lines, then you must start at one of these points and finish at the other. If there are more than two intersections that have an odd number of lines, then you cannot traverse the circuit. The explaination comes down to the fact that for every intersection, you must have a line of entry and a line of escape unless you start from that point or finish at that point. So, you need an even number of lines at a point if you want to go there and leave there.
What is the difference between an Euler circuit and a Hamilton circuit?
An Euler circuit visits each edge once. A Hamiltonian circuit visits each vertex once. It is easy to determine whether an Euler circuit for a given graph exists. It is hard to determine whether a Hamiltonian circuit for a given graph exists.
Hamilton & Euler Circuits???
Think of the degrees of each vertex. There are: 2 vertices of degree 5 5 vertices of degree 2 However, there is a well known theorem stating that to have an Eulerian circuit, a graph must have ALL EVEN degrees. Two of these are odd, which means there is no Eulerian circuit. Why does it have to be that way? Well, in order to move through a vertex, you have to come in on an unused edge, and leave on an unused edge. You can't repeat edges! So every time you come to a vertex, you establish TWO edges on it. That means it has even degree. That same theorem tells us that for Eulerian paths, only the beginning and ending points can have odd degree. That means to have an Eulerian path, all but two of the vertices must have even degrees. That means that this is possible. Based on this, we have to start and end on the part with two vertices (on top in this picture). http://math.colgate.edu/~kellen/interspa... So in summary: Eulerian circuit? No. Eulerian path? Yes. A Hamiltonian path does not exist on most complete bipartite graphs. Why? Because you have to go back and forth between sides. For example, let's call the vertices on one side A,B,C,D,E. The other side can be X,Y. You have to eliminate them one-by-one, but you must alternate from side-to-side. So start with A, then X, then B, then Y, then C... and you're stuck. You can't get to D or E because you need to go through X or Y to do so. The graph K(m,n) has a Hamiltonian path if and only if |m-n|<3.
Help with Euler Circuit problem!?
Euler paths and circuits are actually pretty easy. In a connected graph, an Euler circuit exists if and only if every vertex has even valency. An Euler path exists if and only if at most two vertices have odd valency*. a) is easy, but it depends whether you allow empty paths. If you do, then just the single vertex (no edges) is an empty Euler path and an empty Euler cycle. b) We can't have both vertices with even valency, so they must be odd, say 1 each. That is, just take the graph with two vertices connected by an edge. c) This time, we need the two vertices to have even valency. Just take the graph with two vertices and two edges connecting them. * Thanks to handshake theorem, it's impossible for exactly one vertex to have odd valency, so it's either 2 or 0 vertices with odd degree. EDIT: Yes, a single vertex with a loop would also work. Still, it depends on your definition of a path whether a path exists in that graph. Don't forget, the edge is a cycle, since it starts and ends at the same vertex, and hence is not a path.
What is an efficient algorithm to find an Eulerian circuit in an undirected graph?
Fleury’s Algorithm (Why does Fleury's algorithm have to return to the starting node?)The Algorithm’s pseudo-code (https://www.math.ku.edu/~jmartin...) [Side 65/66]It traverses the Graph just like any traversal, so one can think of it as requiring an Adjacency List. To know which edges are Bridge-Edges we just need to make sure that [math]E(G)[/math]\[math]\{e\}[/math] , for some [math]e[/math] = [math](v_{i},v_{j})[/math] [math]\in[/math] [math]E(G)[/math] does not make G disconnected. This can be done by doing a separate BFS/DFS on the vertices of [math]e[/math] when we are choosing whether or not to remove [math]e[/math].(If [math]\exists [/math][math]v[/math] [math]\in[/math] [math]V(G)[/math] , [math]\ni [/math], deg(v) = [math]2m + 1[/math] for some [math]m[/math] [math]\in[/math] [math]\mathbb{N}[/math]) , then there can never be an Euler Circuit E [math]\in[/math] [math]G[/math]. In other words, [math]\forall[/math] [math]v[/math] [math]\in[/math] [math]V(G)[/math] , [math]v[/math] should have an even degree.
What is the class of the graphs in which every Eulerian circuit is also a Hamilton circuit?
Cycle graphs. Path graphs have the corresponding property for Eulerian and Hamiltonian paths.
One application of Euler circuits is the checking of parking meters. List other real-life?
The basic premise of an Euler circuit is that you have a set of pathways (lined with parking meters, in your example) that need to be traversed. Each pathway begins and ends at a node (a streetcorner, in your example). The goal is to find a way that visits each pathway once, and only once. It's hard work emptying parking meters, and you want to avoid doing any more work than you have to :) You can visit each node as many times as you want, but can only walk down each pathway once. You also need to end up where you started. So, a similar example would be a gardener in a public park. Say there are numerous pathways in the park, and each pathway has a row of flowers along it. The gardener must water all the flowers once and only once. Another example could be painting lines on a new roadway. Say there is a city where all the roads are brand-new and still have yet to be painted. The painters need to paint a stripe down each road, and only one stripe. So they need to find a way that they can go down each new road once and only once (they don't want to paint any more than is absolutely necessary). Hope this helps!