Complete even graph K (3,3) and full graph K5 whether there is a planar representation

This paper discusses the existence of K (3, 3) and K5 plane representation. First, the definition of the plane representation of the graph is given:

If you can draw a graph in the plane and let the edges have no intersections (the intersection of the edges is the line or arc of the edges that intersect outside their public endpoints), the graph is planar. Such a drawing is called the plane representation of the graph.

Obviously, proving that a graph is non-planar is more difficult than proving that a graph is planar. Because for the latter we can use the structural proof of existence to illustrate that a graph is planar.

The first consideration is whether K (3, 3) is planar in nature. in order to solve this problem, we may first think that there is a plane representation, so we begin to try various possibilities, and attempt to use the proof of the existence of the tectonic to find a legal solution. Unfortunately, after trying a lot of possibilities, we still haven't found a legal solution. Naturally, we begin to deny the previous view in our hearts, and instead think that it is impossible to have a flat representation. However, this is just a kind of speculation, standing in the G. polya angle we may say that the discovery of mathematics can not be separated from conjecture, but conjecture is merely conjecture, unproven conjecture is unreliable, the attitude to conjecture should either prove it or overthrow it, Conjecture that can neither be overturned nor proved can become a worldwide problem, such as the famous Goldbach conjecture and the four-color theorem proved by computer. pull away, the ideological thing or directly read Polya book it. Back to our topic, below we will prove that K (3, 3) is non-planar.

Consider two sets, each with two elements (vertices), and two elements in each of the two sets, making a planar representation of the concept of a complete even graph. This is obviously easy to do, it is obviously a quadrilateral, and the vertices belonging to the same set are on the diagonal of the quadrilateral (generalized here, because the quadrilateral may be irregular). Considering that there are still two elements not added, we do not lose the general one of them, the placement of the position is obviously two, either in the quadrilateral area or outside the quadrilateral area, if within the region, and another set of two vertices connected, so that the plane divided into three regions, Notice that the rest of the vertices are not likely to intersect in any area where such a situation is decentralized. The same is the case outside the region. So far we have proved that K (3, 3) is non-planar.

Using similar ideas, we can prove that the complete graph K5 is also non-planar (hits: You can place four vertices, its form must be, consider the fifth vertex, it can not be placed anywhere in the intersection, it is possible to place the first three vertices, it constitutes a triangle, Then consider the placement method of the other two vertices.

Complete even graph K (3,3) and full graph K5 whether there is a planar representation