In Exercises 5–9 determine whether the given graph is planar. If so, draw it so that no edges cr

In Exercises 5–9 determine whether the given graph is...

Key Concepts

Planar Graph

A planar graph is one that can be drawn on a two-dimensional plane without any of its edges crossing each other except at their endpoints. This concept is fundamental in graph theory because it relates to the inherent structure of the graph and whether it can be represented in a way that avoids edge intersections.

Planar Embedding

Planar embedding refers to the specific drawing of a graph in the plane where no edges cross. It involves finding an arrangement of the vertices and edges in such a way that the graph maintains its connectivity without any intersections aside from shared vertices. This is often the goal when determining the planarity of a given graph.

Kuratowski's Theorem

Kuratowski's theorem is a key result in graph theory which states that a finite graph is nonplanar if and only if it contains a subgraph that is a subdivision of either the complete graph K5 or the complete bipartite graph K3,3. This theorem provides a criterion for testing a graph’s planarity by checking for these specific forbidden subgraphs.

Euler's Formula for Planar Graphs

Euler’s formula establishes a relationship among the number of vertices (V), edges (E), and faces (F) in any connected planar graph, given by V - E + F = 2. This formula is used to analyze the structure of planar graphs and can provide necessary conditions for planarity, especially when combined with additional constraints on the graph's structure.

Transcript

00:01 We're asking to determine whether the given graph is pointer and it's so to draw it so no edges cross.

00:10 So we're giving a graph on six vertices.

00:14 They have a, b, c, d, e, and f.

00:34 And we have edges a, c, a, d, a, e, a, f, as well as edges b, c, b, d, and d, f.

01:03 Edges c e and cf, edges d -f, and edges d -f.

01:36 Now notice that given graph doesn't have a sub -graph any more get to k -5.

02:05 And to see why, this is because two of the six vertices in this graph have a degree of three.

02:16 So notice that b has a degree of three.

02:19 D also has a degree of 3.

02:25 E has a degree of 3.

02:31 And we know that k5 requires five vertices of degree of 4 each.

03:40 Now, notice that the given graph also does not have a subgraph homeomorphic 2 to complete bipartite graph k33.

04:07 And this is because there are not three vertices that are connected to all remaining three vertices in the graph.

04:18 So, for example, say we pick vertices is a, b, and c.

04:28 Well, we see that a is connected to f, e, and b.

04:35 B is only connected to f and d, so we can't have that.

04:42 Notice that c also was only connected to f and e.

04:49 So you see that if you choose any three vertices, you won't be able to find any that are connected to all their main three vertices in the graph...

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