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# lect_38 - Planar Graphs II Margaret M Fleck 30 April 2010...

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Planar Graphs II Margaret M. Fleck 30 April 2010 This lecture continues the discussion of planar graphs (section 9.7 of Rosen). 1 Announcements Makeup quiz last day of classes (at the start of class). Your room for the final exam (Friday the 7th, 7-10pm) is based on the first letter of your last name: A-H Roger Adams Lab 116 I-W Business Instructional Facility (515 E. Gregory opposite Armory) 1001 X-Z Roger Adams Lab 116 The conflict exam (Monday the 10th, 1:30-4:30pm) is in 269 Everett Lab. 2 A corollary of Euler’s formula Suppose G is a connected simple planar graph, with v vertices, e edges, and f faces, where v 3. Then e 3 v - 6. 1

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Proof: The sum of the degrees of the regions is equal to twice the number of edges. But each region must have degree 3. So we have 2 e 3 f . Then 2 3 e f . Euler’s formula says that v - e + f = 2, so f = e - v +2. Combining this with 2 3 e f , we get e - v + 2 2 3 e So e 3 - v + 2 0. So e 3 v - 2. Therefore e 3 v - 6. We can also use this formula to show that the graph K 5 isn’t planar. K 5 has five vertices and 10 edges. This isn’t consistent with the formula e 3 v - 6. Unfortunately, this trick doesn’t work for K 3 , 3 , which isn’t planar but satisfies the equation (with 6 vertices and 9 edges). 3 Another corollary In a similar way, we can show that if G is a connected planar simple graph with e edges and v vertices, with v 3, and if G has no circuits of length 3, then e 2 v - 4. Proof: The sum of the degrees of the regions is equal to twice the number of edges. But each region must have degree 4 because we have no circuits of length 3. So we have 2 e 4 f . Then 1 2 e f .
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