In a Hamiltonian path you may not pass through all edges. Graph Theory Definitions In descending order of generality Walk : a sequence of edges where the end of one edge marks the beginning of the next edge Trail : a walk which does not repeat any edges.
All trails are walks. Creating buffer circle x kilometers from point using Python? Can some one tell me the difference between hamiltonian path and euler path. They seem similar! Guy Coder So the term Euler Path or Euler Cycle seems misleading to me.
It should be Euler Trail or Euler Circuit. Abu Nafee Ibna Zahid. I agree with Md. Abu Nafee. Its original name is Eulerian trail. Euler path is a misnomer. Add a comment. Active Oldest Votes. In an Euler path you might pass through a vertex more than once. In a Hamiltonian path you may not pass through all edges. Peter 1 1 gold badge 3 3 silver badges 17 17 bronze badges.
Chris Diver Chris Diver Yes, I believe there are certain properties of a Euler path which you can use to prove a graph has a Euler path without an algorithm to traverse it. Finding a Hamiltonian path is an NP-complete, i think the algorithm involves trial and error.
I thought this would be beyond the scope of the original question to add it to the answer, the OP is obviously new to graph theory :D It's been a while for me, I might dig out my old books. Show 1 more comment. Graph Theory Definitions In descending order of generality Walk : a sequence of edges where the end of one edge marks the beginning of the next edge Trail : a walk which does not repeat any edges.
All trails are walks. Luc 23 3 3 bronze badges. Luckily, Euler solved the question of whether or not an Euler path or circuit will exist. In the graph below, vertices A and C have degree 4, since there are 4 edges leading into each vertex.
B is degree 2, D is degree 3, and E is degree 1. Is there an Euler circuit on the housing development lawn inspector graph we created earlier in the chapter? All the highlighted vertices have odd degree. Since there are more than two vertices with odd degree, there are no Euler paths or Euler circuits on this graph. Unfortunately our lawn inspector will need to do some backtracking.
When it snows in the same housing development, the snowplow has to plow both sides of every street. This can be visualized in the graph by drawing two edges for each street, representing the two sides of the street. Notice that every vertex in this graph has even degree, so this graph does have an Euler circuit. The following video gives more examples of how to determine an Euler path, and an Euler Circuit for a graph.
Not every graph has an Euler path or circuit, yet our lawn inspector still needs to do her inspections. Her goal is to minimize the amount of walking she has to do.
In order to do that, she will have to duplicate some edges in the graph until an Euler circuit exists. Eulerization is the process of adding edges to a graph to create an Euler circuit on a graph. To eulerize a graph, edges are duplicated to connect pairs of vertices with odd degree. Connecting two odd degree vertices increases the degree of each, giving them both even degree.
When two odd degree vertices are not directly connected, we can duplicate all edges in a path connecting the two. For the rectangular graph shown, three possible eulerizations are shown. Notice in each of these cases the vertices that started with odd degrees have even degrees after eulerization, allowing for an Euler circuit.
If we were eulerizing the graph to find a walking path, we would want the eulerization with minimal duplications. If the edges had weights representing distances or costs, then we would want to select the eulerization with the minimal total added weight. Looking again at the graph for our lawn inspector from Examples 1 and 8, the vertices with odd degree are shown highlighted.
With eight vertices, we will always have to duplicate at least four edges. In this case, we need to duplicate five edges since two odd degree vertices are not directly connected.
The problem of finding the optimal eulerization is called the Chinese Postman Problem, a name given by an American in honor of the Chinese mathematician Mei-Ko Kwan who first studied the problem in while trying to find optimal delivery routes for postal carriers. This problem is important in determining efficient routes for garbage trucks, school buses, parking meter checkers, street sweepers, and more. Unfortunately, algorithms to solve this problem are fairly complex.
Some simpler cases are considered in the exercises. In the last section, we considered optimizing a walking route for a postal carrier. How is this different than the requirements of a package delivery driver? While the postal carrier needed to walk down every street edge to deliver the mail, the package delivery driver instead needs to visit every one of a set of delivery locations. Instead of looking for a circuit that covers every edge once, the package deliverer is interested in a circuit that visits every vertex once.
A Hamiltonian circuit is a circuit that visits every vertex once with no repeats. A Hamiltonian path also visits every vertex once with no repeats, but does not have to start and end at the same vertex.
One Hamiltonian circuit is shown on the graph below. There are several other Hamiltonian circuits possible on this graph. Notice that the circuit only has to visit every vertex once; it does not need to use every edge. Notice that the same circuit could be written in reverse order, or starting and ending at a different vertex. Unlike with Euler circuits, there is no nice theorem that allows us to instantly determine whether or not a Hamiltonian circuit exists for all graphs.
We can see that once we travel to vertex E there is no way to leave without returning to C, so there is no possibility of a Hamiltonian circuit. With Hamiltonian circuits, our focus will not be on existence, but on the question of optimization; given a graph where the edges have weights, can we find the optimal Hamiltonian circuit; the one with lowest total weight.
This problem is called the Traveling salesman problem TSP because the question can be framed like this: Suppose a salesman needs to give sales pitches in four cities. He looks up the airfares between each city, and puts the costs in a graph.
In what order should he travel to visit each city once then return home with the lowest cost? To answer this question of how to find the lowest cost Hamiltonian circuit, we will consider some possible approaches. The first option that might come to mind is to just try all different possible circuits. To apply the Brute force algorithm, we list all possible Hamiltonian circuits and calculate their weight:. Since we are summing over the faces, each edge of the polyhedron is counted twice in this sum.
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