Slide 1

1 Discrete Math Graphs, Planar graphs Graph coloring Slide 2 Planar graph A graph is called planar if it can be drawn in the plane without any edges crossing. Such a drawing of a graph, with no crossings, is called a planar representation of the graph A crossing is the intersection of the lines or arcs representing the edges at a point other than a common endpoint 2 Slide 3 Is the graph planar? Can we draw this graph in a plane without any crossings? If so how? 3 a1 a5a4 a2a3 e4 e7 e6 e5e2 e3 e1 Slide 4 First move a5 to the left to remove the crossings, This removes crossings e5-e3, e5-e6, e4-e1, e4-e6, e4-e2 Is the graph planar? 4 a1 a5 a4 a2a3 e5 e4 e7 e6 e3 e1 e2 Slide 5 Then move a4 to the right to remove the remaining crossing e2-e6 Is the graph planar? 5 a1 a5 a4 a2a3 e5 e4 e7 e6 e3 e1 e2 Slide 6 Your turn: 6 B D C E A G F Is this a planar graph Slide 7 Your turn: solution 1 7 B D C E A G F Is this a planar graph B D C E A G F Slide 8 Your turn: solution 2 8 B D C E A G F Is this a planar graph B D C E A G F Slide 9 Eulers formula Theorem: Let G be a connected planar simple graph with e edges and v vertices. Let r be the number of regions in a planar represetnation of G. Then r = e v + 2 Proof by inductive construction of a sequence of subgraphs. Begin with a single edge, add one more edge each step 9 Slide 10 Eulers formula: proof (1) Pick an arbitrary edge of G and the vertices at each end of that arbitrary edge to be our first subgraph G 1 Obtain graph G n from graph G n-1 by arbitrarily picking an edge incident with a vertex that is part of G n-1, but is not itself part of G n-1 and adding it to G n-1 Adding the vertex at the other end of the edge we just added to G n-1 to G n-1 10 Slide 11 Eulers formula: proof (2) Now prove by induction BASE CASE: For an arbitrary edge and the vertices it is incident upon e=1 v=2 so r1=1- 2+2=1. Cleary the plane is a single region surrounding this graph INDUCTIVE HYPOTHESIS: Eulers formula r n =e n -v n +2 holds for n=p INDUCTIVE STEP: show Eulers formula holds for n=p+1, beginning with the inductive hypothesis 11 Slide 12 Eulers formula: proof (3) If the selected edge is incident upon a second node that is already a part of G n. Then the new edge must be inside one of the existing regions 12 The number of edges increases by 1 The number of nodes does not change. The new edge splits one region of the plane into two regions (otherwise lines representing edges would cross. So the number of regions increases by one. e n +1-v n +2=e n+1 -v n+1 +2=r n+1 Slide 13 Eulers formula: proof (3) If the selected edge is incident upon a second node that will be added to G n. 13 The number of edges increases by 1 The number of nodes increases by 1 No new regions are created e n +1-(v n +1)+2=e n+1 -v n+1 +2=r n+1 Using the principle of mathematical induction we have established that Eulers formula holds Slide 14 Example Suppose that a connected simple planar graph has 25 vertices and 12 edges of degree 3 and 13 edges of degree 4. How many regions does this graph divide the plane into? The graph has 0.5(4*13+3*12 )=(52+36)/2=88/2=44 edges So 44-25+2=21 regions are created 14 Slide 15 Corollary If G is a connected planar simple graph with e edges and v vertices, where v3, then e 3v- 6 15 Slide 16 Corollary If G is a connected planar simple graph, the G has a vertex of degree not exceeding five Proof: If there is one vertex or two vertices. A simple graph will be a vertex or an edge containing two vertices. In both of these cases the degree of the vertices cannot exceed 5. If there are at least 3 vertices, lets prove by contradiction 16 Slide 17 Corollary (proof continued) If G is a connected planar simple graph, the G has a vertex of degree not exceeding five If there are at least 3 vertices, lets prove by contradiction we know e3v-6 from previous corallary. So 2e6v-12. Assume the degree of every vertex is at least 6, then by the handshake theorem 2e= deg(v)6v. But this contradicts 2e6v-12 so there must be a vertex with degree