Step 2: Find the next cheapest link of the graph and mark it in blue. The next cheapest link is between B and C with a weight of two miles. Step 3: Find the next cheapest link of the graph and mark it in blue provided it does not make a circuit or it is not a third edge coming out of a single vertex.
The next cheapest link is between D and E with a weight of three miles. Step 4: Find the next cheapest link of the graph and mark it in blue provided it does not make a circuit or it is not a third edge coming out of a single vertex.
The next cheapest link is between A and E with a weight of four miles, but it would be a third edge coming out of a single vertex. The next cheapest link is between A and C with a weight of five miles. Mark it in blue. Step 5: Since all vertices have been visited, close the circuit with edge DA to get back to the home office, A.
This is the only edge we could close the circuit with because AB creates three edges coming out of vertex B and BD also created three edges coming out of vertex B. Efficient Algorithm: an algorithm for which the number of steps needed to carry it out grows in proportion to the size of the input to the problem. Approximate Algorithm: any algorithm for which the number of steps needed to carry it out grows in proportion to the size of the input to the problem.
There is no known algorithm that is efficient and produces the optimal solution. Reference Point : the starting point of a Hamilton circuit. Hamilton Circuit. A Hamiltonian path visits every node or vertex exactly once, and a Eulerian path traverses every edge exactly once.
They are related but are neither dependent nor mutually exclusive. As a result, vertices can be repeated but edges cannot. K5 has 5! The edges consist of both the red lines and the dotted black lines. The red lines show a Hamiltonian circuit that this graph contains. If you start at any node, and follow the red lines, you will touch each node exactly once before you arrive back at your starting node. So by definition, this is a Hamiltonian graph. A path is a sequence of vertices with the property that each vertex in the sequence is adjacent to the vertex next to it.
A path that does not repeat vertices is called a simple path. A circuit is path that begins and ends at the same vertex. An Euler path is a path that travels through all edges of a connected graph. Proving a graph has no Hamiltonian cycle [closed] A graph with a vertex of degree one cannot have a Hamilton circuit. Moreover, if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit.
A Hamilton circuit cannot contain a smaller circuit within it. Definition: A graph is considered Hamiltonian if and only if the graph has a cycle containing all of the vertices of the graph. Definition: A Hamiltonian cycle is a cycle that contains all vertices in a graph. 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:.
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 visit 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 of 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. In other words, heuristic algorithms are fast, but are not guaranteed to produce the optimal circuit. Consider our earlier graph from Example 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 a salesperson visitng the five cities shown in the graph below. From there:. Going back to Example One option would be to redo the nearest neighbor algorithm with a different starting point to see if the result changed.
Since nearest neighbor is so fast, doing it several times isn't a big deal. We will revisit the graph from Example The RNNA was able to produce a slightly better circuit with a weight of 25, but still not the optimal circuit in this case.
The table below shows the time, in milliseconds, it takes to send a packet of data between computers on a network. If data needed to be sent in sequence to each computer, then notification needed to come back to the original computer, we would be solving the TSP.
The computers are labeled A-F for convenience. Section
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