# What is a hamiltonian path in a graph

Hamiltonian path

Hamiltonian Path and Hamiltonian Circuit- Hamiltonian path is a path in a connected graph that contains all the vertices of the graph. A closed Hamiltonian path is . Hamiltonian Path is a path in a directed or undirected graph that visits each vertex exactly once. The problem to check whether a graph (directed or undirected) contains a Hamiltonian Path is NP-complete, so is the problem of finding all the Hamiltonian Paths in a graph. Following images explains the idea behind Hamiltonian Path more clearly.

In the next lesson, we will investigate specific kinds of paths how to program comcast remote to receiver a graph called Euler paths and circuits.

Euler paths are an optimal path through a graph. They are named after him because it was Euler who first defined them. By counting the number of vertices of a graph, and their degree we can determine whether a graph has an Euler path or circuit. We will also learn another algorithm that will allow us to find an Euler circuit once we determine that a graph has one.

Because Euler first studied this question, these types of paths are named after him. An Euler path is a path that uses every edge in a graph with no repeats. Being a path, it does not have to return to the starting vertex. In the graph shown below, there are several Euler paths. The path is shown in arrows to the right, with the order of edges numbered.

An Euler circuit is a circuit that uses every edge in a graph with no repeats. Being a circuit, it must start and end at the same vertex. The graph below has several possible Euler circuits. The second is shown in arrows. Look back at the example used for Euler paths—does that graph have an Euler circuit? A few tries will tell you no; that graph does not have an Euler circuit.

When we were working with shortest paths, we were interested in the optimal path. Why do we care if an Euler circuit exists? Think back to our housing development lawn inspector from the beginning of the chapter. The lawn inspector is interested in walking as little as possible. The ideal situation would be a circuit that covers every street with no repeats.

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 how to get aimbot for black ops ps3 online 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 how to work out wattage 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 how to build a chicken wire garden fence calculate their weight:.

Note: These are the unique circuits on this graph. All other possible circuits are the reverse of the listed ones or start at a different vertex, but result in the same weights. The Brute force algorithm is optimal; it will always produce the Hamiltonian circuit with minimum weight.

Is it efficient? To answer that question, we need to consider how many Hamiltonian circuits a graph could have. This is called a complete graph.

Suppose we had a complete graph with five vertices like the air travel graph above. From Seattle there are four cities we can what phones from straight talk use verizon first. From each of those, there are three choices. From each of those cities, there are two possible cities to visit next. There is then only one choice for the last city before returning home.

For N vertices in a complete graph, there will be [latex] n-1! Half how to instal tamil fonts for windows 7 the circuits are duplicates of other circuits but in reverse order, leaving unique routes. But consider what happens as the number of cities increase:.

As you can see the number of circuits is growing extremely quickly. If a computer looked at one billion circuits a second, it would still take almost two years to examine all the possible circuits with only 20 cities! Certainly Brute Force is not an efficient algorithm. Unfortunately, no one has yet found an efficient and optimal algorithm to solve the TSP, and it is very unlikely anyone ever will.

Since it is not practical to use brute force to solve the problem, we turn instead to heuristic algorithms ; efficient algorithms that give approximate solutions. In other words, heuristic algorithms are fast, but may or may not produce the optimal circuit. We ended up finding the worst circuit in the graph! What happened? Unfortunately, while it is very easy to implement, the NNA is a greedy algorithm, meaning it only looks at the immediate decision without considering the consequences in the future.

In this case, following the edge AD forced us to use the very expensive edge BC later. Consider again our salesman. From there:. Going back to our first example, how could we improve the outcome?

One option would be to redo the nearest neighbor algorithm with a different starting point to see if the result changed. This is the same circuit we found starting at vertex A. No better.

Aug 23,  · Hamiltonian graph - A connected graph G is called Hamiltonian graph if there is a cycle which includes every vertex of G and the cycle is called Hamiltonian cycle. Hamiltonian walk in graph G is a walk that passes through each vertex exactly once. Dirac's Theorem - If G is a simple graph with n vertices, where n ? 3 If deg(v) ? {n}/{2} for each vertex v, then the graph G is Hamiltonian graph. Sep 27,  · Hamilonian Circuit – A simple circuit in a graph that passes through every vertex exactly once is called a Hamiltonian circuit. Unlike Euler paths and circuits, there is no simple necessary and sufficient criteria to determine if there are any Hamiltonian paths or circuits in a graph. Jun 09,  · A Hamiltonian path is a path that visits each vertex of the graph exactly once. A Hamiltonian path can exist both in a directed and undirected graph. It’s important to discuss the definition of a path in this scope: It’s a sequence of edges and vertices in which all the vertices are rutlib6.com: Subham Datta.

Join Stack Overflow to learn, share knowledge, and build your career. Connect and share knowledge within a single location that is structured and easy to search. The problem of testing whether a graph G contains a Hamiltonian path is NP-hard , where a Hamiltonian path P is a path that visits each vertex exactly once.

There does not have to be an edge in G from the ending vertex to the starting vertex of P , unlike in the Hamiltonian cycle problem. I am planning to use DFS and Topological sorting. But I didn't know how to connect the two concepts in solving the problem. How can a topological sort be used to determine the solution. Once this is done, you know that edge go from lower index vertices to higher.

This means that there exists a Hamiltonian path if and only if there are edge between consecutive vertices, e. This is because in a Hamiltonian path you can't "go back" and yet you have to visit all, so the only way is to "not skip". Stack Overflow for Teams — Collaborate and share knowledge with a private group.

Create a free Team What is Teams? Learn more. Asked 8 years ago. Active 3 years, 3 months ago. Viewed 29k times. I am referring to Skienna's Book on Algorithms. Any suggestions? Improve this question. Add a comment. Active Oldest Votes.

This is because in a Hamiltonian path you can't "go back" and yet you have to visit all, so the only way is to "not skip" You can check this condition in O n. Improve this answer. Petar Ivanov Petar Ivanov But you assumed the graph is connected, can't there be a topological sort for a graph that has disconnected parts?