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.
How to find an Euler path/circuit with 6 vertices?
Search engine. Next time, try that first. You're asking people to needlessly duplicate efforts when you ask general knowledge questions. So which do you want? Path or circuit? A simple hexagon is a 6 point Euler path. I won't describe a 6 point circuit... so look one up.
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.
Is it possible to have both Euler path and Euler Circuit? explain please?
http://www.youtube.com/watch?v=REfC1-igKHQ the above youtube link explains the differences There are different criteria for a Euler path and a Euler circuit Euler path....even number of ways from each vertex except for 2 which are odd Euler circuit...all number of ways from each vertex are even Click on the blue line above for a good explaining demonstration
Determine the value(s) of n for which the complete graph Kn has an Euler circuit.?
a) For odd values of n, each vertex of K_n has even degree so K_n has an Euler circuit. b) The graph K_2 has an Euler path but not an Euler circuit. For even values of n greater than 2, K_n has more than 2 vertices of odd degree, so K_n has neither an Euler path nor an Euler circuit. See "Properties" on Wikipedia and "Eulerian Path" on MathWorld.
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.
Euler Path and Circuit equation? Help!?
What is D? D is one of a number of special functions related to circuits in complete graphs. The D(x, y) function provides a convenient means of representing and evaluating the number of Hamilton and Euler circuits in a complete graph Kn . I expect you have already found that for complete graph Kn with even n there are no Euler circuits. Why is there a comma? It takes the form D(x, y) so the comma separates those D(x, y) = y!*{(2x + 1)!/[(x)! *2^x]}^y I think you copied that formula incorrectly. It is simpler For n odd, the number of Euler circuits E(n) in complete graph Kn is given by E(n) = (n - 1)*D[(n - 3)/2, n - 2)] For example when n = 5 E(5) = 4D[1, 3] = 4*3!*{3!/2]^3 = 6^4/2 = 648 This Algana Associates web page gives a brief summary of such functions http://www.algana.co.uk/publications/Cir... Regards - Ian