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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 K3,3, can K3,3 be drawn in the plane so that no two of its edges cross? K3,3
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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. R4 R R2 R1
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Regions In any planar representation of K3,3, vertex v1 must be connected to both v4 and v5, and v2 also must be connected to both v4 and v5. v1 v v3 v4 v v6
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Regions The four edges {v1, v4}, {v4, v2}, {v2, v5}, {v5, v1} form a closed curve that splits the plane into two regions, R1 and R2. v v5 R2 R1 v v2
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Regions Next, we note that v3 must be in either R1 or R2.
Assume v3 is in R2. Then the edges {v3, v4} and {v4, v5} separate R2 into two subregions, R21 and R22. v v v1 v5 R21 R2 R1 → v3 R22 v v v v2
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Regions Now there is no way to place vertex v6 without forcing a crossing: If v6 is in R1 then {v6, v3} must cross an edge If v6 is in R21 then {v6, v2} must cross an edge If v6 is in R22 then {v6, v1} must cross an edge v v5 R21 v R1 R22 v v2
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Regions Alternatively, assume v3 is in R1. Then the edges {v3, v4} and {v4, v5} separate R1 into two subregions, R11 and R12. v v5 R11 R R12 v3 v v2
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Regions Now there is no way to place vertex v6 without forcing a crossing: If v6 is in R2 then {v6, v3} must cross an edge If v6 is in R11 then {v6, v2} must cross an edge If v6 is in R12 then {v6, v1} must cross an edge v v5 R11 R R12 v3 v v2
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Planar Graphs Consequently, the graph K3,3 must be nonplanar. K3,3
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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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Euler’s 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. R R R2 R1 # of edges, e = 6 # of vertices, v = 4 # of regions, r = e - v + 2 = 4
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Euler’s 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 K5 planar? K5
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Euler’s Formula (Cont.)
K5 has 5 vertices and 10 edges. We see that v 3. So, if K5 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, K5 is nonplanar. K5
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Euler’s 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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Euler’s 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 K3,3 planar?
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Euler’s Formula (Cont.)
K3,3 has 6 vertices and 9 edges. Obviously, v 3 and there are no circuits of length 3. If K3,3 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 K3,3 is nonplanar. K3,3
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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 A B C D E F G c b a d f g e b a e d c
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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 What is the chromatic number of the graph shown below?
The chromatic number must be at least 3 since a, b, and c must be assigned different colors. So Let’s try 3 colors first. 3 colors work, so the chromatic number of this graph is 3. b e a d g c f
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Example What is the chromatic number for each of the following graphs?
White White Yellow Yellow Green Yellow White White Yellow Yellow White Chromatic number: 2 Chromatic number: 3
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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? Assume that in all other pairs there are students. L25
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Graph Coloring and Schedules
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: 3203 3261 3137 4115 1007 4118 3157 4156 L25
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Graph Coloring and Schedules
One way to do this is to put edges down where students mutually excluded… 3203 3261 3137 4115 1007 4118 3157 4156 L25
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Graph Coloring and Schedules
…and then compute the complementary graph: 3203 3261 3137 4115 1007 4118 3157 4156 L25
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Graph Coloring and Schedules
…and then compute the complementary graph: 3203 3261 3137 4115 1007 4118 3157 4156 L25
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Graph Coloring and Schedules
Redraw: 3137 3203 3261 4115 1007 3157 4118 4156 L25
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Graph Coloring and Schedules
Not 1-colorable because of edge 3137 3203 3261 4115 1007 3157 4118 4156 L25
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Graph Coloring and Schedules
Not 2-colorable because of triangle 3137 3203 3261 4115 1007 3157 4118 4156 L25
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Graph Coloring and Schedules
Is 3-colorable. Try to color by Red, Green, Blue. 3137 3203 3261 4115 1007 3157 This method is also would have discovered if graph were not 3-colorable and given a proof by contradiction. 4118 4156 L25
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Graph Coloring and Schedules
WLOG Red, 3157-Blue, 4118-Green: 3137 3203 3261 4115 1007 3157 4118 4156 L25
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Graph Coloring and Schedules
So 4156 must be Blue: 3137 3203 3261 4115 1007 3157 4118 4156 L25
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Graph Coloring and Schedules
So 3261 and 4115 must be Red. 3137 3203 3261 4115 1007 3157 4118 4156 L25
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Graph Coloring and Schedules
3137 and 1007 easy to color. 3137 3203 3261 4115 1007 3157 4118 4156 L25
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Graph Coloring and Schedules
So need 3 exam slots: Slot 2 3137 3203 3261 Slot 1 4115 1007 3157 4118 Slot 3 4156 L25
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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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