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CSE 211 Discrete Mathematics and Its Applications Chapter 8.7 Planar Graphs.

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Presentation on theme: "CSE 211 Discrete Mathematics and Its Applications Chapter 8.7 Planar Graphs."— Presentation transcript:

1 CSE 211 Discrete Mathematics and Its Applications Chapter 8.7 Planar Graphs

2 The House-and-Utilities Problem

3 Planar Graphs Consider the previous slide. Is it possible to join the three houses to the three utilities in such a way that none of the connections cross?

4 Planar Graphs Phrased another way, this question is equivalent to: Given the complete bipartite graph K 3,3, can K 3,3 be drawn in the plane so that no two of its edges cross? K 3,3

5 Planar Graphs A graph is called planar if it can be drawn in the plane without any edges crossing. A crossing of edges is the intersection of the lines or arcs representing them at a point other than their common endpoint. Such a drawing is called a planar representation of the graph.

6 Example A graph may be planar even if it is usually drawn with crossings, since it may be possible to draw it in another way without crossings.

7 Example A graph may be planar even if it represents a 3-dimensional object.

8 Planar Graphs We can prove that a particular graph is planar by showing how it can be drawn without any crossings. However, not all graphs are planar. It may be difficult to show that a graph is nonplanar. We would have to show that there is no way to draw the graph without any edges crossing.

9 Regions Euler showed that all planar representations of a graph split the plane into the same number of regions, including an unbounded region. R 4 R 3 R 2 R 1

10 Regions In any planar representation of K 3,3, vertex v 1 must be connected to both v 4 and v 5, and v 2 also must be connected to both v 4 and v 5. v 1 v 2 v 3 v 4 v 5 v 6

11 Regions The four edges {v 1, v 4 }, {v 4, v 2 }, {v 2, v 5 }, {v 5, v 1 } form a closed curve that splits the plane into two regions, R 1 and R 2. v 1 v 5 R 2 R 1 v 4 v 2

12 Regions Next, we note that v 3 must be in either R 1 or R 2. Assume v 3 is in R 2. Then the edges {v 3, v 4 } and {v 4, v 5 } separate R 2 into two subregions, R 21 and R 22. v 1 v 5 v 1 v 5 R 21 R 2 R 1 v 3 R 22 v 4 v 2 v 4 v 2

13 Regions Now there is no way to place vertex v 6 without forcing a crossing: –If v 6 is in R 1 then {v 6, v 3 } must cross an edge –If v 6 is in R 21 then {v 6, v 2 } must cross an edge –If v 6 is in R 22 then {v 6, v 1 } must cross an edge v 1 v 5 R 21 v 3 R 1 R 22 v 4 v 2

14 Regions Alternatively, assume v 3 is in R 1. Then the edges {v 3, v 4 } and {v 4, v 5 } separate R 1 into two subregions, R 11 and R 12. v 1 v 5 R 11 R 2 R 12 v 3 v 4 v 2

15 Regions Now there is no way to place vertex v 6 without forcing a crossing: –If v 6 is in R 2 then {v 6, v 3 } must cross an edge –If v 6 is in R 11 then {v 6, v 2 } must cross an edge –If v 6 is in R 12 then {v 6, v 1 } must cross an edge v 1 v 5 R 11 R 2 R 12 v 3 v 4 v 2

16 Planar Graphs Consequently, the graph K 3,3 must be nonplanar. K 3,3

17 Regions Euler devised a formula for expressing the relationship between the number of vertices, edges, and regions of a planar graph. These may help us determine if a graph can be planar or not.

18 Eulers Formula Let G be a connected planar simple graph with e edges and v vertices. Let r be the number of regions in a planar representation of G. Then r = e - v + 2. # of edges, e = 6 # of vertices, v = 4 # of regions, r = e - v + 2 = 4 R 4 R 3 R 2 R 1

19 Eulers Formula (Cont.) Corollary 1: If G is a connected planar simple graph with e edges and v vertices where v 3, then e 3v - 6. Is K 5 planar? K5K5

20 Eulers Formula (Cont.) K 5 has 5 vertices and 10 edges. We see that v 3. So, if K 5 is planar, it must be true that e 3v – 6. 3v – 6 = 3*5 – 6 = 15 – 6 = 9. So e must be 9. But e = 10. So, K 5 is nonplanar. K5K5

21 Eulers Formula (Cont.) Corollary 2: If G is a connected planar simple graph, then G has a vertex of degree not exceeding 5.

22 Eulers Formula (Cont.) Corollary 3: If a connected planar simple graph has e edges and v vertices with v 3 and no circuits of length 3, then e 2v - 4. Is K 3,3 planar?

23 Eulers Formula (Cont.) K 3,3 has 6 vertices and 9 edges. Obviously, v 3 and there are no circuits of length 3. If K 3,3 were planar, then e 2v – 4 would have to be true. 2v – 4 = 2*6 – 4 = 8 So e must be 8. But e = 9. So K 3,3 is nonplanar. K 3,3

24 CSE 211 Discrete Mathematics Chapter 8.8 Graph Coloring

25 Introduction When a map is colored, two regions with a common border are customarily assigned different colors. We want to use the smallest number of colors instead of just assigning every region its own color.

26 4-Color Map Theorem It can be shown that any two-dimensional map can be painted using four colors in such a way that adjacent regions (meaning those which sharing a common boundary segment, and not just a point) are different colors.

27 Map Coloring Four colors are sufficient to color a map of the contiguous United States. Source of map: http://www.math.gatech.edu/~thomas/FC/fourcolor.html

28 Dual Graph Each map in a plane can be represented by a graph. –Each region is represented by a vertex. –Edges connect to vertices if the regions represented by these vertices have a common border. –Two regions that touch at only one point are not considered adjacent. The resulting graph is called the dual graph of the map.

29 Graph Theoretic Foundations Dual graph:

30 Dual Graph Examples A B C D E F G A B C D E b a e d c c b a d f g e

31 Graph Coloring A coloring of a simple graph is the assignment of a color to each vertex of the graph so that no 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 the graph. The Four Color Theorem: The chromatic number of a planar graph is no greater than four.

32 Example The chromatic number must be at least 3 since a, b, and c must be assigned different colors. So Lets try 3 colors first. 3 colors work, so the chromatic number of this graph is 3. What is the chromatic number of the graph shown below? b e a d g c f

33 Example What is the chromatic number for each of the following graphs? White Yellow White Yellow White Yellow Chromatic number: 2 Chromatic number: 3 White Yellow White Yellow Green

34 L2534 Graph Coloring and Schedules EG: Suppose want to schedule some final exams for CS courses with following call numbers: 1007, 3137, 3157, 3203, 3261, 4115, 4118, 4156 Suppose also that there are no common students in the following pairs of courses because of prerequisites: 1007-3137 1007-3157, 3137-3157 1007-3203 1007-3261, 3137-3261, 3203-3261 1007-4115, 3137-4115, 3203-4115, 3261-4115 1007-4118, 3137-4118 1007-4156, 3137-4156, 3157-4156 How many exam slots are necessary to schedule exams?

35 L2535 Turn this into a graph coloring problem. Vertices are courses, and edges are courses which cannot be scheduled simultaneously because of possible students in common: 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

36 L2536 One way to do this is to put edges down where students mutually excluded… 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

37 L2537 …and then compute the complementary graph: 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

38 L2538 …and then compute the complementary graph: 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

39 L2539 Redraw: 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

40 L2540 Not 1-colorable because of edge 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

41 L2541 Not 2-colorable because of triangle 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

42 L2542 Is 3-colorable. Try to color by Red, Green, Blue. 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

43 L2543 WLOG. 3203-Red, 3157-Blue, 4118-Green: 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

44 L2544 So 4156 must be Blue: 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

45 L2545 So 3261 and 4115 must be Red. 1007 3137 3157 3203 4115 3261 4156 4118 Graph Coloring and Schedules

46 L2546 Graph Coloring and Schedules 3137 and 1007 easy to color. 1007 3137 3157 3203 4115 3261 4156 4118

47 L2547 So need 3 exam slots: 1007 3137 3157 3203 4115 3261 4156 4118 Slot 1 Slot 2 Slot 3 Graph Coloring and Schedules

48 Conclusion In this chapter we have covered: –Introduction to Graphs –Graph Terminology –Representing Graphs and Graph Isomorphism –Graph Connectivity –Euler and Hamilton Paths –Planar Graphs –Graph Coloring


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