Im using the equation. Is that what you mean here? If not, disregard the rest of this answer. For a Hamilton circuit, you need to visit every vertex once. To simplify things, let's pick a particular vertex and think of it as the start vertex for all of our circuits. I call it v1. To build a Hamilton circuit, you need to choose the next vertex to visit.
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 a graph has a Hamiltonian cycle, then the graph is said to be Hamiltonian. A Hamilton path in a graph is a path that includes each vertex of the graph once and only once. A Hamilton circuit is a circuit that includes each vertex of the graph once and only once.
At the end, of course, the circuit must return to the starting vertex. Definition: A graph is considered Eulerian if the graph is both connected and has a closed trail a walk with no repeated edges containing all edges of the graph. The graph on the left is Eulerian. Cheapest Link Algorithm Pick an edge with the cheapest weight, in case of a tie, pick whichever pleases you. Colour your edge. Pick the next cheapest uncoloured edge unless: your new edge closes a smaller circuit. Repeat Step 2 until the hamilton circuit is complete.
A Hamiltonian path is a simple open path that contains each vertex in a graph exactly once. The Hamiltonian Path problem is the problem to determine whether a given graph contains a Hamiltonian path. In graph theory, a path in a graph is a finite or infinite sequence of edges which joins a sequence of vertices which, by most definitions, are all distinct and since the vertices are distinct, so are the edges. A complete graph is a graph that has an edge between every single vertex in the graph ; we represent a complete graph with n vertices using the symbol Kn.
Therefore, the first example is the complete graph K7, and the second example isn't a complete graph at all. A tournament with more than two vertices is Hamiltonian if and only if it is strongly connected. Dirac's theorem on chordal graphs, the characterization of chordal graphs as graphs in which all minimal separators are cliques. Depth first search and backtracking can also help to check whether a Hamiltonian path exists in a graph or not. Simply apply depth first search starting from every vertex v and do labeling of all the vertices.
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. 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. Notice that this is actually the same circuit we found starting at C, just written with a different starting vertex. 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. The next shortest edge is BD, so we add that edge to the graph. Notice that the algorithm did not produce the optimal circuit in this case; the optimal circuit is ACDBA with weight While the Sorted Edge algorithm overcomes some of the shortcomings of NNA, it is still only a heuristic algorithm, and does not guarantee the optimal circuit.
0コメント