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This is also known as the vertex 1-center problem and can be extended to the vertex k-center problem. Finding the center of a graph is useful in facility location problems where the goal is to minimize the worst-case distance to the facility. For example, placing a hospital at a central point reduces the longest distance the ambulance has to ...
Vertex distance. Vertex distance is the distance between the back surface of a corrective lens, i.e. glasses (spectacles) or contact lenses, and the front of the cornea. Increasing or decreasing the vertex distance changes the optical properties of the system, by moving the focal point forward or backward, effectively changing the power of the ...
In graph theory, the metric k-center problem or vertex k-center problem is a classical combinatorial optimization problem studied in theoretical computer science that is NP-hard. Given n cities with specified distances, one wants to build k warehouses in different cities and minimize the maximum distance of a city to a warehouse.
For example, to find the midpoint of the path, substitute σ = 1 ⁄ 2 (σ 01 + σ 02); alternatively to find the point a distance d from the starting point, take σ = σ 01 + d/R. Likewise, the vertex, the point on the great circle with greatest latitude, is found by substituting σ = + 1 ⁄ 2 π. It may be convenient to parameterize the ...
The latter may occur even if the distance in the other direction between the same two vertices is defined. In the mathematical field of graph theory, the distance between two vertices in a graph is the number of edges in a shortest path (also called a graph geodesic) connecting them. This is also known as the geodesic distance or shortest-path ...
[15] In the set cover problem formed from a metric dimension problem, the elements to be covered are the () pairs of vertices to be distinguished, and the sets that can cover them are the sets of pairs that can be distinguished by a single chosen vertex. The approximation bound then follows by applying standard approximation algorithms for set ...
As noted above, the iterative solution to the inverse problem fails to converge or converges slowly for nearly antipodal points. An example of slow convergence is (Φ 1, L 1) = (0°, 0°) and (Φ 2, L 2) = (0.5°, 179.5°) for the WGS84 ellipsoid. This requires about 130 iterations to give a result accurate to 1 mm. Depending on how the inverse ...
The idea is to reduce this problem to a network flow problem. Let G′ = (V′ = A ∪ B, E′ = E). Assign the capacity of all the edges in E′ to 1. Add a source vertex s and connect it to all the vertices in A′ and add a sink vertex t and connect all vertices inside group B′ to this vertex. The capacity of all the new edges is 1 and ...