Applied Combinatorics, 4 th Ed. Alan Tucker

Applied Combinatorics, 4th Ed.Alan Tucker Section 1.4 Planar Graphs Prepared by Michele Fretta and Jessica Scheld

a b f d c Planar Graphs A graph is planar if it can be drawn on a plane without edges crossing.A plane graph is a planar depiction of a planar graph. Plane graph of Figure 1 Figure 1 a e e c d b f

Example 1: Map Coloring • Question: How many different colors are needed to color the countries on a map so that any pair of countries with a common border have different colors? • A map is a planar graph, with edges as borders and vertices where borders meet. • Dual graph: obtained by making a vertex for each country and an edge between vertices corresponding to two countries with a common border

a b c d a b c d Dual Graph Example

Map coloring E G A B We need 4 colors to color the map without any bordering countries having the same color. D C F It’s easier to see it in the second representation of the graph, since it shows that A,B,C,D are a separate subgraph from D,E,F,G. Thus, we would need no more than 4 colors. Note that this dual graph does not include an outer region.

Circle-Chord Method(one method of determining if a graph is planar) • What we want is to prove that some graph is planar (no edges cross). • The circle-chord method: • Step 1: Find a circuit that contains all the vertices of our graph (draw it as a large circle) • Note: Finding a circuit which includes all vertices is difficult. • Recall: circuit is a path that ends where it began • Step 2: The remaining non-circuit edges, called chords, must be drawn either inside or outside the circle in a planar drawing.

Step 1 a b c d e f g h Find a circuit. The orange highlights the circuit on this graph. a f e c Draw this circuit as a circle. b h g d

a f e c b h g d Step 2 a b c d e f g h Step 2: Choose a chord to draw, either inside or outside the circle. For example, we will start with the chord (b,f), drawing it outside the circle. Since no lines cross another, this graph is planar!

Showing is Nonplanar • indicates that the graph is a complete bipartite graph consisting of two sets of 3 vertices with each vertex in one set adjacent to all vertices in the other set. 1 3 5 2 4 6 1 3 5 2 4 6 1 2 6 5 3 4

1 2 6 5 3 4 Configuation • Note: 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. • Finding the configuration can be tricky.

Example of a configuration 1 A configuration proves a graph is nonplanar. It has 6 vertices of degree 3, plus some number of vertices of degree 2 that subdivide the edges of it. Here those vertices are 3 and 6. 2 8 3 7 4 6 5 In order to find a configuration, eliminate vertices of degree 2 that subdivide edges of it. The remaining graph looks like the depiction of earlier.

1 2 3 1 2 5 4 6 5 3 4 Theorem 1 (Kuratowski, 1930) • A graph is planar if and only if it does not contain a subgraph that is or configuration. • Recall: is a graph on n vertices with an edge joining every pair of vertices.( is a triangle.) is a bipartite graph with m and n vertices in its two vertex sets and all possible edges between vertices in the two sets.

Theorem 2: Euler’s Formula • If G is a connected planar graph, then any plane graph depiction of G has r = e - v+ 2 regions. • Recall: Connected planar graphs have paths between each pair of vertices. • v = number of vertices • e= number of edges • r= number of regions • This is important because there are many different plane graph depictions that can be drawn for a planar graph, however, the number of regions will not change.

Proof of Euler’s Formula • Let’s draw a plane graph depiction of G, edge by edge. • Let denote the connected plane graph after n edges have been added. • Let denote the number of vertices in • Let denote the number of edges in • Let denote the number of regions in

Proof of Euler’s Formula (cont’d) • Let’s start by drawing Euler’s formula is valid for , since r=e-v+2 1=1-2+2 We obtain from by adding an edge at one of the vertices of .

Proof of Euler’s Formula (cont’d) In general, we can obtain from by adding an edge to one of the vertices of . • The new edge might link two vertices already in . • Or, the new edge might add another vertex to . We will use the method of induction to complete the proof: We have shown that the theorem is true for . Next, let’s assume that it is true for for any n>1, and prove it is true for . Let (x,y) be the edge that is added to to get . There are two cases to consider.

y K x Proof of Euler’s Formula (cont’d) In the first case, x and y are both in . Then they are on the boundary of a common region K, possibly an unbounded region. Edge (x, y) splits K into two regions. Then, Each side of Euler’s formula grows by one. So, if the formula was true for , it will also be true for .

x y K Proof of Euler’s Formula (cont’d) In the second case, one of the vertices x,y is not in . Let’s say that it is x. Then, adding (x,y) implies that x is also added, but that no new regions are formed (no existing regions are split). So, the value on each side of Euler’s equation is unchanged. The validity of Euler’s formula for implies its validity for . By induction, Euler’s formula is true for all ’s and the full graph G.

Example of Euler’s Formula • How many regions would there be in a plane graph with 10 vertices each of degree 3?

Corollary • If G is a connected planar graph with e > 1, then e ≤ 3v – 6 • Proof: • Define the degree of a region as the number of edges on its boundary. If an edge occurs twice along the boundary, then count it twice. The region K has degree 12. K

Proof of Corollary continued • Note that no region can be less than degree 3. • A region of degree 2 would be bounded by two edges joining the same pair of vertices (parallel edges) • A region of degree 1 would be bounded by a loop edge. • Neither of these is allowed, and so a region must have at least degree 3.

Proof of Corollary continued

Example: Show is nonplanar. • Corollary: If G is a connected planar graph with e > 1, then e ≤ 3v – 6 • Use the Corollary. has 5 vertices and 10 edges. Thus, we have 3v – 6 = 3 x 5 – 6 = 9. But, e ≤ 3v – 6 must be true in a connected planar graph, thus, cannot be planar.

b f a e c d Exercise:Is this graph planar? Step 1: Draw a circuit a – b – e – d – c – f – a

b f a e c d Exercise (cont’d) Step 2: Draw the circuit as a circle. Step 3: Add the remaining edges. a f b c e d

Applied Combinatorics, 4 th Ed. Alan Tucker