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9.7 Planar Graphs

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Intro problem- 3 houses and 3 utilities K 3,3 problem: Can 3 houses be connected to 3 utilities so that no 2 lines cross? Similarly, can an isomorphic version of K 3,3 be drawn in the plane so that no two edges cross? TXY houses UVW utilities

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Planar Def Def: A graph is called planar if it can be drawn in the plane without any edges drawing.

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Sketchpad examples Check examples-- See Fig01 for K 4, K 5, K 2,3, K 3,3

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Q3, Q4 Check examples— Fig02 for Q 3, Q 4

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Sketch05

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Sketch06

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Sketch07

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Proof that K 3,3 is not planar see Fig 01 and Sketch 08 and Math Teacher article

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Proof Consider a subgraphT2 U1V X

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…proof Pf. 1: Case 1: W is outside the graph (region 2). This forms region 2a and 2b. Y must be adjacent to U, V, W…. T UVW X

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Case 2 Case 2: W is inside the graph (region 1). This forms regions 1a and 1b Again, Y is adjacent to U, V, and W…. T UWV X

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Is K5 planar? (see Fig01)

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Claim: K 5 is nonplanar. Proof: By contradiction… Suppose there is a planar representation of K5. 2 So v1, v2, v3, v4, v5 form a pentagon. 1 3 5 4 {v1,v3} must be present. WLOG, let it be on the inside. Then construct {v2,v4} and {v2,v5} on the outside. So __________ are on the __________

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Find # of regions, edges, vertices to discover Euler’s formula rev W3W3 K 2,3 Q3Q3

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Theorem 1: Euler’s Formula Thm: 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 = _______ Proof: First, specify a planar representation of G. We will prove by specifying a sequence of subgroups G 1, G 2, … G e =G, adding an edge at each step. This is possible because G is connected. Arbitrarily pick an edge of G to obtain G 1. Obtain G n from G n-1 by arbitrarily adding an edge that is incident with a vertex in G n-1, adding the other vertex if necessary.

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…proof outline By induction: Basis: e=1 G 1 r 1 = ___ e 1 = ___ v 1 = ___ So_________ Inductive step: Assume n and show n+1. This means: Assume r n = e n – v n + 2 and add {a n+1, b n+1 } to G n to obtain G n+1 and show ___________… Case 1: a n+1, b n+1 G n R is split into 2 regoins. r n+1 = ___ e n+1 = ___ v n+1 = ___ So ___________ Case 2: a n+1 G n but b n+1 G n r n+1 = ___ e n+1 = ___ v n+1 = ___

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Question: How do you prove a graph is either planar or not planar? To prove it is… To prove it isn’t…

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Corollary 1: e≤3v-6 Corollary 1: If G is a connected planar simple graph with e edges and v vertices where v≥3, then e≤3v-6. Def: deg(R )= number of edges on the boundary of region R Proof: Assume G is simple. Therefore it has no loops or multiple edges. So it has no regions of degree 1 or 2. A planar representation of G has r regions, each of degree at least 3. Note: 2e = ≥ 3r So r ≤ (2/3)e Using Euler’s Theorerm, r = ______ ≤ _____ …

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Ex: Use the contrapositive of Corollary 1 to prove that K 5 is nonplanar.

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Q: Can you use Corollary 1 to show K3,3 is nonplanar?

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Cor. 2– region degree ≤ 5 A Corollary of Cor. 1 is the following: Cor. 2: If G is a connected planar simple graph, then G has a vertex of degree not exceeding 5. Proof: Case 1: G has 1 or 2 vertices: result _______ Case 2: G has at least 3 vertices. By ___, we know e ≤ ____ so 2e ≤ ____ To show result, assume degree of every vertex is ______. Then because 2e = _____ by _______, we have 2e ≥ 6v (why?). But this contradicts ___. So there must be a vertex with degree ≤ 5.

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Corollary 3: e ≤ 2v – 4 Corollary 3: If G is a connected planar simple graph with e edges, v vertices, v≥3, and no circuits of length 3, then e≤2v – 4. Proof. Assume G is simple. Consider a planar representative of G. Therefore it has no loops or multiple edges, which would create regions of degree 1 or 2. With no circuits of length 3, there are no regions of degree 3. Therefore, all regions are at least degree 4. So 2e = ________ ≥ 4r Solving for r… By Euler’s Formula…

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K3,3 Ex: Use Corollary 2 to prove that K3,3 is nonplanar.

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Is Q4 planar or not? Prove.

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Kuratowski’s Theory Def: Replacing {u,v} with {u,w} and {w,v} is an elementary subdivision. Def: G1=(V1,E1) and G2=(V2,E2) are homeomorphic if they can be obtained from the same graph by a sequence of elementary subdivisions. Kuratowski’s Theorem: A graph is nonplanar iff it contains a subgraph homeomorphic to K3,3 or K5. Proof: clear beyond scope of class

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examples Are the following planar or not? Why?

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Peterson ex Ex: Use the two Euler Corollaries on the Peterson example (See examples in notes) Cor. 1: e ≤ 3v - 6 Cor. 3: e ≤ 2v - 4

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Kuratowski’s Theory Ex: Use Kuratowski’s Theory on the Peterson example. see written handout

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More ex See handout for more examples using Euler and Kuratowski – See sketch06 and sketch07 on sketchpad to the right – See attached Sketchpad handout with 7 more ex

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More – handout ex 1-2

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Handout 3-4

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Handout 5-7

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Ex- book 7, 23

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