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CSE 211 Discrete Mathematics and Its Applications Chapter 8.7 Planar Graphs

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The House-and-Utilities Problem

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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?

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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

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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.

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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.

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

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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.

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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

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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

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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

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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

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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

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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

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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

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

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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.

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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

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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

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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

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

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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?

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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

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CSE 211 Discrete Mathematics Chapter 8.8 Graph Coloring

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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.

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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.

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Map Coloring Four colors are sufficient to color a map of the contiguous United States. Source of map:

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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.

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Graph Theoretic Foundations Dual graph:

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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

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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.

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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

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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

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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: , , , , , , , , , How many exam slots are necessary to schedule exams?

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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: Graph Coloring and Schedules

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L2536 One way to do this is to put edges down where students mutually excluded… Graph Coloring and Schedules

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L2537 …and then compute the complementary graph: Graph Coloring and Schedules

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L2538 …and then compute the complementary graph: Graph Coloring and Schedules

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L2539 Redraw: Graph Coloring and Schedules

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L2540 Not 1-colorable because of edge Graph Coloring and Schedules

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L2541 Not 2-colorable because of triangle Graph Coloring and Schedules

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L2542 Is 3-colorable. Try to color by Red, Green, Blue Graph Coloring and Schedules

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L2543 WLOG Red, 3157-Blue, 4118-Green: Graph Coloring and Schedules

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L2544 So 4156 must be Blue: Graph Coloring and Schedules

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L2545 So 3261 and 4115 must be Red Graph Coloring and Schedules

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L2546 Graph Coloring and Schedules 3137 and 1007 easy to color

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L2547 So need 3 exam slots: Slot 1 Slot 2 Slot 3 Graph Coloring and Schedules

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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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