Suppose there are three cottages on a plane (or sphere) and each needs to be connected to the gas, water, and electric companies. Using a third dimension or sending any of the connections through another company or cottage are disallowed. Is there a way to make all nine connections without any of the lines crossing each other?

Solution

The answer is no; It is impossible to connect the three cottages with the three different utilities without at least one of the connections crossing another.

As You Can The Possible Graph Made to fulfill All the Condition is Like This

The problem is part of the mathematical field of topological graph theory which studies the embedding of graphs on surfaces. In more formal graph-theoretic terms, the problem asks whether the complete bipartite graph K3,3 is planar. This graph is often referred to as the utility graph in reference to the problem.[1] The graph is equivalent to the circulant graph Ci6(1,3). Kazimierz Kuratowski proved in 1930 that K3,3 is nonplanar, and thus that the problem has no solution.

One proof of the impossibility of finding a planar embedding of K3,3 uses a case analysis involving the Jordan curve theorem, in which one examines different possibilities for the locations of the vertices with respect to the 4-cycles of the graph and shows that they are all inconsistent with a planar embedding. Alternatively, it is possible to show that any bridgeless bipartite planar graph with n vertices and m edges has m ≤ 2n − 4 by combining the Euler formula n - m + f = 2 (where f is the number of faces of a planar embedding) with the observation that the number of faces is at most half the number of edges (because each face has at least four edges and each edge belongs to exactly two faces). In the utility graph, m = 9 and 2n − 4 = 8, violating this inequality, so the utility graph cannot be planar.

Two important characterizations of planar graphs, Kuratowski's theorem that the planar graphs are exactly the graphs that contain neither K3,3 nor the complete graph K5 as a subdivision, and Wagner's theorem that the planar graphs are exactly the graphs that contain neither K3,3 nor K5 as a minor, encompass this result.

## No comments:

Post a Comment