Tucker Applied Combinatorics Section 1 4 prepared by

Tucker, Applied Combinatorics, Section 1. 4, prepared by Patti Bodkin Planar Graphs Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 1

A graph is called planar if it can be drawn on a plane without edges crossing. A plane graph is a drawing of planar graph in the plane. a a b e e f c b d d This is a plane graph of Figure 1. f c Figure 1 Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 2

Here are two ways to determine if a graph is planar: The first is a practical heuristic, which the author refers to as the circle-chord method. It consists of a step-by-step method of drawing the graph, edge-by-edge without crossing any edges. The second consists of theoretical results, such as Kuratowski’s Theorem, or Euler’s Formula. Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 3

Circle-Chord Method--Planar Step One: Find a circuit that contains all the vertices of the graph. (Recall: a circuit is a path that ends where it starts) **Note: step one is not always possible, and is often difficult** a b c d e f g h Jan. 27, 2003 a Tucker, Applied Combinatorics, Sect. 1. 4 a The circuit for this figure is highlighted in red 4

Circle-Chord Method--Planar Step Two: Draw this circuit as a large circle. a a b c d e f b c e f g h h g d Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 5

Circle-Chord Method--Planar Step Three: Choose one chord, and decide to draw it either inside or outside the circle. If chosen correctly, it will force certain other chords to be drawn opposite to the circle. (Inside if the first chord was drawn outside, and vice versa. ) a b c a d e f b c e f g h h Since the chords could be drawn without crossing, this graph is planar. Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 g d 6

Circle-Chord Method--Nonplanar Step One: Find a circuit that contains all the vertices of the graph. (Recall: a circuit is a path that ends where it starts) **Note: step one is not always possible, and is often difficult** a d The circuit for Figure 2 is highlighted in red b h c Figure 2 e f g Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 7

Circle-Chord Method--Nonplanar Step Two: Draw this circuit as a large circle. a a h d b b h c e c g f g d f e The remaining edges (in black) must be drawn either inside or outside the circle. These edges are called the chords. Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 8

Circle-Chord Method--Nonplanar Step Three: Choose one chord, and decide to draw it either inside or outside the circle. If chosen correctly, it will force certain other chords to be drawn opposite to the circle. (Inside if the first chord was drawn outside, and vice versa. ) a h a d b c g b h c d f e f Because these lines cross, this is not the plane graph for Figure 2. However, this does not mean that the graph is not planar. Often, it is very difficult to find the correct planar graph. e g Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 9

Recall: and configurations. We say that a subgraph is a configuration if it can be obtained from a by adding vertices in the middle of some edges. A configuration is defined similarly. Here a has been subdivided by adding vertices in the middle of some of the edges. Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 10

Kuratowski’s Theorem A graph is planar if and only if it does not contain a subgraph that is a or configuration. The configuration is more common in nonplanar graphs than . If the graph is nonplanar than it contains one of the configurations with Left Side added vertices. a a h b b c g d f e Jan. 27, 2003 H and F were removed to show the configuration c g Tucker, Applied Combinatorics, Sect. 1. 4 Right Side d e 11

A graph is defined to be a connected planar graph if it’s in “one-piece”, i. e. there are paths between every pair of vertices. Euler’s Formula Assume that is a connected planar graph, with: V = # of vertices E = # of edges R = # of regions. Euler’s Formula states that if is a connected planar graph, then any plane graph depiction of has regions. Figure 3: a b e d Jan. 27, 2003 c Tucker, Applied Combinatorics, Sect. 1. 4 12

Euler’s Formula a a c b a e b e c d c a b e d Jan. 27, 2003 c Tucker, Applied Combinatorics, Sect. 1. 4 13

Proof of Euler’s Formula Draw a plane graph depiction of edge-by-edge. Let denote the (connected) plane graph obtained after n edges have been added, and let , , , denote the number of vertices, edges and regions in respectively. Initially there is , which consists of one edge, it’s two end vertices and the one (unbounded) region. Then , a , , and so Euler’s formula is valid for since , c We obtain Jan. 27, 2003 from by adding an edge at one of the vertices in Tucker, Applied Combinatorics, Sect. 1. 4 . 14

Proof of Euler’s Formula (cont. ) In general, vertices of is obtained from. , by adding an The new edge might link two vertices already in If it does not, the other end vertex of the added to. edge at one of the. edge is a new vertex that must be Now we will use the method of induction to complete the proof. We have shown that theorem is true for Next we assume that it is true for Let be the . for any edge that is added to , and prove it is true for to get . . There are two cases to consider… Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 15

Proof of Euler’s Formula (cont. ) In the first case, and are both in. Then they are on the boundary of a common region unbounded region. of , possibly the See Figure 4 Edge splits into two regions. Then, Figure 4 So each side of Euler’s formula grows by 1. Hence if the formula was true for , it will also be true for Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 . 16

Proof of Euler’s Formula (cont. ) In the second case, one of the vertices is not in say it is. Then adding implies that is also added, but no new regions are formed (no existing regions are split). See Figure 5 Thus formula is unchanged. and the value on each side of Euler’s The validity of Euler’s formula for induction, the formula is true for all Jan. 27, 2003 implies its validity for and for the full graph Tucker, Applied Combinatorics, Sect. 1. 4 . By. 17

Worked Example Book Example # 5 How many regions would there be in a plane graph with 10 vertices each of degree 3? (Recall: the degree of a vertex is the number of edges incident to the vertex) By theorem in Section 1. 3, the sum of the degrees, 10 x 3, equals = 15. By Euler’s formula, the number of regions is: Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 , and so 18

Corollary to Euler’s Formula (Thm. 2) If is a connected planar graph with , then . Proof: The degree of a region is defined to be the number of edges on its boundary. To be precise, we say that if an edge occurs twice along a boundary, as does in region in Figure 6, the edge is counted twice in region degree; for example, region has degree 10 and region has degree 3 in the figure. Figure 6 Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 19

Each region in a plane graph must have degree , for a region of degree 2 would be bounded by 2 edges joining the same pair of vertices and a region of degree 1 would be bounded by a loop edge, but parallel edges and loops are not allowed in graphs (in this book. ) Since each region has degree , the sum of degrees of all regions will be at least. But this sum of degrees of all regions must equal , since this sum counts each edge twice. That is each of an edge’s two sides is part of some boundary. Thus, = (sum of regions’ degrees) Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 20

Corollary to Euler’s Formula (Thm. 2) Proof: (cont. ) Combining this inequality with Euler’s formula we have: Figure 4: Solving for e, we obtain: In this case, edge is counted twice in the deg( ). The deg( ) is 3. Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 21

CAREFUL! Do not interpret the corollary as meaning: If , then a connected graph is planar, because there are many nonplanar graphs which also satisfy this equation. For example, has 6 vertices and 9 edges. So when you substitute into the equation, you get: , which holds. However, is not planar. Jan. 27, 2003 Tucker, Applied Combinatorics, Sect. 1. 4 22

FOR THE CLASS TO TRY: Use the Corollary to Euler’s Formula to prove whether the following configuration is planar. Substituting into the Corollary: Configuration However, the corollary states that connected planar graph, and so the Jan. 27, 2003 must be true in a configuration cannot be planar. Tucker, Applied Combinatorics, Sect. 1. 4 23