Chapter 10 7 Planar Graphs These class notes

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Chapter 10. 7 Planar Graphs These class notes are based on material from our

Chapter 10. 7 Planar Graphs These class notes are based on material from our textbook, Discrete Mathematics and Its Applications, 7 th ed. , by Kenneth H. Rosen, published by Mc. Graw Hill, Boston, MA, 2011. 1

Planar Graphs • Consider the next slide. Is it possible to join the three

Planar Graphs • Consider the next slide. Is it possible to join the three houses to the three utilities in such a way that none of the connections cross? 2

The House-and-Utilities Problem 3

The House-and-Utilities Problem 3

Planar Graphs • Phrased another way, this question is equivalent to: Given the complete

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 4

Planar Graphs • A graph is called planar if it can be drawn in

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. 5

Example A graph may be planar even if it is usually drawn with crossings,

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. 6

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

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

Planar Graphs • We can prove that a particular graph is planar by showing

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. 8

Regions • Euler showed that all planar representations of a graph split the plane

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 9

Regions • Euler devised a formula for expressing the relationship between the number of

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. 10

Euler’s Formula • Let G be a connected planar simple graph with e edges

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 4 R 3 R 2 R 1 # of edges, e = 6 # of vertices, v = 4 # of regions, r = e - v + 2 = 4 11

Euler’s Formula Example • r=e-v+2 12

Euler’s Formula Example • r=e-v+2 12

Euler’s Formula (Cont. ) • r=e-v+2 • Corollary 1: If G is a connected

Euler’s Formula (Cont. ) • r=e-v+2 • Corollary 1: If G is a connected planar simple graph with e edges and v vertices where v 3, then e 3 v - 6. (no proof) • Is K 5 planar? K 5 13

Euler’s Formula (Cont. ) • • Corollary 1: e ≤ 3 v – 6

Euler’s Formula (Cont. ) • • Corollary 1: e ≤ 3 v – 6 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 3 v – 6 = 3*5 – 6 = 15 – 6 = 9. So e must be 9. But e = (n choose 2) = n(n-1)/2 = 5*4/2 = 10. • So, K 5 is nonplanar. K 5 14

Euler’s Formula (Cont. ) • Corollary 2: If G is a connected planar simple

Euler’s Formula (Cont. ) • Corollary 2: If G is a connected planar simple graph, then G must have a vertex of degree not exceeding 5. (no proof) • 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 2 v - 4. (no proof) 15

Euler’s Formula (Cont. ) • Is K 3, 3 planar? 16

Euler’s Formula (Cont. ) • Is K 3, 3 planar? 16

Euler’s Formula (Cont. ) • • Corollary 3: … e 2 v - 4

Euler’s Formula (Cont. ) • • Corollary 3: … e 2 v - 4 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 2 v – 4 would have to be true. 2 v – 4 = 2*6 – 4 = 8 So e must be 8. But e = 9. So K 3, 3 is nonplanar. K 3, 3 17

Kuratowski Theorem A graph is nonplanar if and only if it contains a subgraph

Kuratowski Theorem A graph is nonplanar if and only if it contains a subgraph homeomorphic to K 3, 3 or K 5. The intuition: homeomorphism is like graph isomorphism after ignoring some of degree 2 nodes. Example: after ignoring nodes d, e, f in graph H, H is isomorphic to K 5. 18

Theorem. Graph K 3, 3 is nonplanar. • Proof (direct proof with simple geometry).

Theorem. Graph K 3, 3 is nonplanar. • Proof (direct proof with simple geometry). 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 19

Regions • The four edges {v 1, v 4}, {v 4, v 2}, {v

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 2 v 3 v 1 v 5 R 2 v 4 v 5 v 6 v 4 R 1 v 2 20

Regions • Next, we note that v 3 must be in either R 1

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 R 21 R 2 R 1 → v 3 R 22 v 4 v 2 21

Regions • Now there is no way to place vertex v 6 without forcing

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 22

Regions • Alternatively, assume v 3 is in R 1. Then the edges {v

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 v 4 R 12 v 3 23

Regions • Now there is no way to place vertex v 6 without forcing

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 v 4 R 12 v 3 24

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

Planar Graphs • Consequently, the graph K 3, 3 must be nonplanar. K 3, 3 25