Planarity

Planar Graphs A graph is called planar if it can be drawn on the plane in a way that no two of its edges cross each other A planar graph? Yes!

Why care? Many practical applications. Nodes: gates Edges: wires between the gates Find a circuit layout w/o intersecting wires

Three houses and three utilities Is K(3,3) planar?

Today’s question Given a graph G What if we can’t? If we can find a drawing of G where no edges cross, then we know that G is planar What if we can’t? Either G is non-planar, or It’s our fault Our goal is to find properties to determine if a graph is planar. For example, we just saw that K(4) is planar. How about K(5)?

Plane graphs A plane graph is a drawing of a planar graph where no edges cross. More common term: planar embedding Not a plane-graph A plane-graph

Regions A plane graph divides the plane into connected pieces called regions Boundary of a region: the subgraph that contains vertices and edges that define the region (those that are incident with the region) R1 R3 R4 R2 R4: exterior (note that it is unbounded)

Euler Identity For any connected plane graph G, n-m+r = 2 where n = order of G(number of vertices) m = size of G (number of edges) r = number of regions of G

Proof Case I: G has no cycles (i.e. a tree) Let n be the number of vertices m = ? r = ? n – (n-1) + 1 = 2 So the theorem holds for trees

Proof (cont) Else (G has a cycle) Assume that the theorem is not true Let G be the smallest size graph where the theorem does not hold G is not a tree G has a cycle (hence a non-bridge edge e) n – m + r  2 Let G’ = G - e The euler identity holds for G’: n – (m-1) + r’ = 2 What is r’?

Theorem If G is a planar graph of order n  3 and size m, m  3n – 6 Interpretation: the number of edges is small. It is O(n) as opposed to O(n^2)

Proof Suppose G is connected Let’s take care of some special cases G has 3 vertices If m = 2, the inequality holds: m = 2  3n – 6 = 3 G is connected and n 3 and m  3 Suppose G has r regions R_1, …, R_r Let m_i = number of edges on the boundary of R_i m_i  3

Proof (cont) Let’s add all the edges region by region In M, every edge appears at most twice, so M  2m Hence we have 3r  M  2m → 3r  2m Apply Euler n - m + r = 2 → 6 = 3n–3m+3r  3n-3m+2m = 3n – m

Proof (cont) So far, we established that for connected planar graphs m  3n – 6 What about disconnected graphs ?

So far We have established our first sufficient condition (for non-planarity). Count the number of edges, if it is more than 3n -6 then non-planar In other words, m  3n – 6 is a necessary condition for planarity (but we don’t know if it is sufficient – i.e. are there non-planar graphs that satisfy this condition) Let’s revisit K(5) n = 5 m = 5*4/2 = 10 3n-6 = 9 So it is not planar

Corollary If G is planar, then G contains a vertex of degree 5 or less. Planar → vertex with degree  5 All vertices have degree  6 → non-planar

(All vertices have degree  6) → non-planar Hence, G is not planar

Let’s go over some of the properties we used again In the sum: An edge appears once if it is a bridge, twice othervise

Back to the three houses and utilities problem: K(3,3) m = 9  3n – 6 = 12 Recall that this condition was necessary for planarity but we still don’t know if it is sufficient. Therefore we can not say whether K(3,3) is planar or not for sure. In fact…

Theorem K(3,3) is non-planar. (hence, our condition is not sufficient) Proof: Assume, to the contrary that, it is. Apply Euler identity n-m+r = 2 → r = 5 Let R1, …, R5 be the five regions and m1,…,m5 be the number of edges on the boundaries as before

Proof (cont) K(3,3) has no triangles, so mi 4 for each Ri So m  10 A contradiction! No bridges

Moral of the story Our sufficient condition is not necessary. But we know that if a graph contains K(5) or K(3,3) as a subgraph then it is not planar. Unfortunately, this is also not a “sufficient-and-necessary” condition There exist non-planar graphs which contain neither K(5) nor K(3,3). See the book for an example.

Subdivision A graph G’ is a subdivision of G if it is obtained by adding one or more vertices of degree 2 into edges of G

Subdivisions Every subdivision of a planar graph is planar Why? Every subdivision of a non-planar graph is also non-planar So if a graph contains a subdivision of K(5) or K(3,3) we know that it is not planar

Theorem (Kuratowski) A graph G is planar if and only if G does not contain K(5) or a subdivision of it K(3,3) or a subdivision of it

Discussion Proof of Kuratowski’s theorem is beyond the scope of our course. Now that we have a sufficient-and-necessary condition, can you come up with an algorithm to test if a graph is planar?