Loading in 5 sec....

Planar Graphs Graph ColoringPowerPoint Presentation

Planar Graphs Graph Coloring

- By
**duena** - Follow User

- 135 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Planar Graphs Graph Coloring' - duena

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### PlanarGraphsGraphColoring

Chapter 9.7-9.8

Introduction

- Consider the complete bipartite graph K3,3 .
- Can K3 ,3 be drawn in the plane so that no two of its edges cross?
- In this section we will study the question of whether a graph can be drawn in the planewithout edges crossing.
- There are always many ways to represent a graph.
- When is it possible to find at least oneway to represent this graph in a plane without any edges crossing?

Planar Graph

- A graph is called planar if it can be drawn in the plane without any edges crossing (wherea crossing of edges is the intersection of the lines or arcs representing them at a point otherthan their common endpoint).
- Such a drawing is called a planar representation of the graph.

Examples

- Is K4 planar?
- Is Q3 planar?

Example

- Is K3,3 planar?
- In any planar representation of K3,3 , the vertices v1 and v2 must be connected to both v4 and v5 . These four edges form a closed curve that splits the plane into two regions, R1 and R2. The vertex v3 is in either R1 or R2 .
- When v3 is in R2 , the inside ofthe closed curve, the edges between v3 and v4 and between v3 and v5 separate R2 into twosubregions, R21 and R22.

Example

- there is no way to place the final vertex v6 without forcing a crossing:
- If v6 is in R1, then the edge between v6 and v3 cannot be drawn without a crossing.
- If v6 is inR21 , then the edge between v2 and v6 cannot be drawn without a crossing.
- If v6 is in R22 , thenthe edge between v1 and v6 cannot be drawn without a crossing.

- A similar argument can be used when v3 is in R1.
- It follows that K3,3 is not planar.

Euler’s Formula

- A planar representation of a graph splits the plane into regions, including an unbounded region.
- For instance, the planar representation of the graph shown in the figure splits the plane into 6 regions.
- Euler showed that all planar representations of a graphsplit the plane into the same number of regions.

Euler’s Formula

- Let G be a connected planar simple graph with e edges and vvertices. Let r be the number of regions in a planar representation of G . Then r = e - v + 2.
- Proof: First, we specify a planar representation of G . We will prove the theorem by constructinga sequence of subgraphs G1, G2 , ..., Ge = G, successively adding an edge at each stage.
- Arbitrarily pick one edge of G to obtain G1 .
- The proof will now proceed by induction. The relationship r1= e1 – v1+ 2 is true for G1 ,because e1 = 1 , v1 = 2, and r1 = 1 .

Euler’s Formula- Proof

- Now assume that rn = en - vn + 2. Let {an+1, bn+1} be the edge that is added to Gn toobtain Gn+1.
- There are two possibilities to consider. In the first case, both an+1 and bn+1 arealready in Gn. These two vertices must be on the boundary of a common region R, or elseit would be impossible to add the edge {an+1, bn+1} to Gn without two edges crossing (andGn+1 is planar). The addition of this new edge splits R into two regions.
- Consequently, in thiscase, rn+1 = rn+1, en+1= en + 1 , and vn+1= vn . So, this formula is still true.
- rn+ 1 = en+1 - vn+1 + 2.

Euler’s Formula- Proof

- In the second case, one of the two vertices of the new edge is not already in Gn . Supposethat an+1 is in Gn but that bn+1 is not. Adding this new edge does not produce any new regions.
- Consequently, rn+1 = rn, en+1 = en+1 and vn+1 = vn+1. So the formula is still true.
- rn+1 = en+1- vn+1+ 2

- We have completed the induction argument. Hence, rn = en - vn + 2 for all n .
- Because theoriginal graph is the graph Ge, obtained after e edges have been added, the theorem is true.

Corollary 1

- If G is a connected planar simple graph with e edges and v vertices, where v 3, thene 3v-6.

Corollary 3

- If a connected planar simple graph has e edges and v vertices with v3 and no circuits oflength three, then e2v - 4.
- Use Corollary 3 to show that K3,3 is nonplanar.
- Because K3,3 has no circuits of length three,Corollary 3 can be used.
- K3,3 has six vertices and nine edges.
- Because e = 9 and 2v - 4 = 8,Corollary 3 shows that K3,3 is nonplanar.

Elementary Subdivision

- If a graph is planar, so will be any graph obtained by removing an edge { u , v} and adding anew vertex w together with edges { u , w } and {w , v} .
- Such an operation is called an elementarysubdivision.
- The graphs G1= ( V1 , E1) and G2 = ( V2 , E2) are called homeomorphic if theycan be obtained from the same graph by a sequence of elementary subdivisions.

Example

- Show that the graphs G1 and G2 displayed in the figure are homeomorphic.
- These graphs are homeomorphic because G2 can be obtained from G1 byelementary subdivisions. G1 can be obtained from itself by an empty sequence of elementarysubdivisions.
- remove the edge {a , c}, add the vertex f, and add the edges {a , f} and {f, c}
- remove theedge {b, c}, add the vertex g, and add the edges {b, g} and {g, c }
- remove the edge{b, g}, add the vertex h , and add the edges {g, h} and {b, h}

Kuratowski's Theorem

- A graph is nonplanar if and only if it contains a subgraph homeomorphic to K3,3 or K5.

Example

- Determine whether the graph G shown in the figure is planar.
- G has a subgraph H homeomorphic to K5.
- H is obtained by deleting h , j , and k andall edges incident with these vertices.
- H is homeomorphic to K5 because it can be obtainedfrom K5 (with vertices a, b, c, g, and i) by a sequence of elementary subdivisions, adding thevertices d, e, and f.
- Hence, G is nonplanar.

Graph Coloring

- Consider the problem of determining theleast number of colors that can be used to color a map so that adjacent regions never havethe same color.
- Each map in the plane can be represented by a graph.
- To set up this correspondence,each region of the map is represented by a vertex.
- Edges connect two vertices if the regionsrepresented by these vertices have a common border.

Definitions

- A coloring of a simple graph is the assignment of a color to each vertex of the graph so thatno two adjacent vertices are assigned the same color.
- The chromatic number of a graph is the least number of colors needed for a coloring of thisgraph. The chromatic number of a graph G is denoted by (G).
- The Four Color Theorem: The chromatic number of a planar graph is no greaterthan four.

Example

- What are the chromatic numbers of the graph G?
- The chromatic number of G is at least three, because the vertices a , b , and c mustbe assigned different colors.
- To see if G can be colored with three colors, assign red to a , blueto b, and green to c.
- Then, d can (and must) be colored red because it is adjacent to b and c.
- Furthermore, e can (and must) be colored green because it is adjacent only to vertices coloredred and blue, and f can (and must) be colored blue because it is adjacent only to vertices coloredred and green.
- Finally, g can (and must) be colored red because it is adjacent only to verticescolored blue and green. This produces a coloring of G using exactly three colors.

Example

- What is the chromatic number of Kn ?
- A coloring of Kn can be constructed using n colors by assigning a different colorto each vertex.
- No two vertices canbe assigned the same color, because every two vertices of this graph are adjacent.
- Hence, thechromatic number of Kn = n .
- Recall that Kn is not planar when n5,

Applications of Graph Colorings

- Scheduling Final Exams:How can the final exams at a university be scheduled so that nostudent has two exams at the same time?
- Supposethere are seven finals to be scheduledandcourses are numbered1 through 7.
- Suppose that the following pairs of courses have common students: 1 and 2,1 and 3, 1 and 4, 1 and 7, 2 and 3, 2 and 4, 2 and 5, 2 and 7, 3 and 4, 3 and 6, 3 and 7, 4 and 5,4 and 6, 5 and 6, 5 and 7, and 6 and 7.
- Because the chromatic number of this graph is 4 four timeslots are needed.

Applications of Graph Colorings

- Frequency Assignments: Television channels 2 through 13 are assigned to stations in NorthAmerica so that no two stations within 150 miles can operate on the same channel.
- How can theassignment of channels be modeled by graph coloring?

Download Presentation

Connecting to Server..