9.7 Planar Graphs
Intro problem- 3 houses and 3 utilities
K3,3 problem: Can 3 houses be connected to 3 utilities so that no 2 lines cross?
Similarly, can an isomorphic version of K3,3 be drawn in the plane so that no two edges cross?
T X Y houses U V W utilities
Sketchpad examples
Check examples--
See Fig01 for K4, K5, K2,3, K3,3
Complete graphs K n
Cycle graphsC n
W heelsW n
Complete bipartite graphs K n,m
…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
U V W X
Case 2
Case 2: W is inside the graph (region 1). This forms regions 1a and 1bAgain, Y is adjacent to U, V, and W….
T
U W V
X
Claim: K5 is nonplanar.
Proof: By contradiction…
Suppose there is a planar representation of K5. 2So 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 __________
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 G1, G2, … Ge
=G, adding an edge at each step. This is possible because G is connected.
Arbitrarily pick an edge of G to obtain G1.
Obtain Gn from Gn-1 by arbitrarily adding an edge that is incident with a vertex in Gn-1, adding the other vertex if necessary.
…proof outlineBy induction: Basis: e=1 G1
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: an+1, bn+1 Gn
R is split into 2 regoins. r n+1 = ___ e n+1 = ___ v n+1 = ___
So ___________Case 2: an+1 Gn but bn+1 Gn
r n+1 = ___ e n+1 = ___ v n+1 = ___
Question: How do you prove a graph is either planar or not planar?
• To prove it is…
• To prove it isn’t…
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)eUsing Euler’s Theorerm, r = ______ ≤ _____
…
Cor. 2– region degree ≤ 5A 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.
Corollary 3: e ≤ 2v – 4Corollary 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 = ________ ≥ 4rSolving for r…By Euler’s Formula…
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:clearbeyond scope of class
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
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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HGFE
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