# When is a graph planar?

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1 When is a graph planar? Theorem(Euler, 1758) If a plane multigraph G with k components has n vertices, e edges, and f faces, then n e + f = 1 + k. Corollary If G is a simple, planar graph with n(g) 3, then e(g) 3n(G) 6. If also G is triangle-free, then e(g) 2n(G) 4. Corollary K 5 and K 3,3 are non-planar. The subdivision of edge e = xy is the replacment of e with a new vertex z and two new edges xz and zy. The graph H is a subdivision of H, if one can obtain H from H by a series of edge subdivisions. Vertices of H with degree at least three are called branch vertices. Theorem(Kuratowski, 1930) A graph G is planar iff G does not contain a subdivision of K 5 or K 3,3. 1

2 Kuratowski s Theorem Theorem(Kuratowski, 1930) A graph G is planar iff G does not contain a subdivision of K 5 or K 3,3. Proof. A Kuratowski subgraph of G is a subgraph of G that is a subdivision of K 5 or K 3,3. A minimal nonplanar graph is a nonplanar graph such that every proper subgraph is planar. A counterexample to Kuratowski s Theorem constitutes a nonplanar graph that does not contain any Kuratowski subgraph. Kuratowski s Theorem follows from the following Main Lemma and Theorem. 2

3 The spine of the proof Main Lemma. If G is a graph with fewest edges among counterexamples, then G is 3-connected. Lemma 1. Every minimal nonplanar graph is 2-connected. Lemma 2. Let S = {x, y} be a separating set of G. If G is a nonplanar graph, then adding the edge xy to some S-lobe of G yields a nonplanar graph. Main Theorem.(Tutte, 1960) If G is a 3-connected graph with no Kuratowski subgraph, then G has a convex embedding in the plane with no three vertices on a line. A convex embedding of a graph is a planar embedding in which each face boundary is a convex polygon. Lemma 3. If G is a 3-connected graph with n(g) 5, then there is an edge e E(G) such that G e is 3-connected. Notation: G e denotes the graph obtained from G after the contraction of edge e. Lemma 4. G has no Kuratowski subgraph G e has no Kuratowski subgraph. 3

4 Proof of Tutte s Theorem Main Theorem. (Tutte, 1960) If G is a 3-connected graph with no Kuratowski subgraph, then G has a convex embedding in the plane with no three vertices on a line. Proof. Induction on n(g). Base case: G is 3-connected, n(g) = 4 K 4. Let e E s.t. H = G e is 3-connected. (Lemma 3) Then H has no Kuratowski subgraph. (Lemma 4) Induction H has a convex embedding in the plane with no three vertices on a line. Let z V (H) be the contracted e. H z is 2-connected boundary of the face containing z after the deletion of the edges incident to z is a cycle C. 4

5 Let x 0,..., x k 1 be the neighbors of x on C in cyclic order. Note that N(x) 3 and hence k 2. Denote by x i, x i+1 the portion of C from x i to x i+1 (including endpoints; indices taken modulo k). Let N x = N(x) \ {y} and N y = N(y) \ {x}. Case 1. N x N y 3. Let u, v, w N x N y. Then x, y, u, v, w are the branch vertices of a K 5 -subdivision in G. Case 2. N x N y 2. Since N x N y 3, there is w.l.o.g. a vertex u N y \ N x. Let i be such that u is on x i, x i+1. Case 2a. N y is contained in x i, x i+1. Then there is an appropriate embedding of G: Placing x in place of z and y sufficiently close to x maintains convexity. (No three vertices are collinear; N(x), N(y) 3.) Case 2b. For every i there is a vertex in N y that is not in x i, x i+1. Then there must be a v N y that is not on x i, x i+1 and x, y, x i, x i+1, u, v are the branch vertices of a K 3,3 - subdivision in G.

6 Proof of the Lemmas Lemma 3. G is 3-connected, n(g) 5 there is an edge e E(G) such that G e is 3-connected. Proof. Suppose G is 3-connected and for every e E, G e is NOT 3-connected. For edge e = xy, the vertex z is the mate of xy if {x, y, z} is a cut in G. Choose e = xy and their mate z such that G {x, y, z} has a component H, whose order is as large as possible. Let H be another component of G {x, y, z} and let u V (H ) be the neighbor of z (There IS one!). Let v be the mate of uz. V (H) {x, y} \ {v} is connected in G {z, u, v} contradicting the maximality of H. 5

7 Lemma 4. G has no Kuratowski subgraph G e has no Kuratowski subgraph. Proof. Suppose G e contains a Kuratowski subgraph H. Then z V (H) z is a branchvertex of H N H (z) = 4 and N H (z) N G (x), N H (z) N G (y) 2 Then H is the subdivision of K 5 G contains a sudivision of K 3,3, a contradiction.

8 Minors K 7 is a toroidal graph (it is embeddable on the torus), K 8 is not. What else is not? For the torus there is NO equivalent version of Kuratowski s characterization with a finite number of forbidden subdivisions. Any such characterization must lead to an infinite list. A weaker concept: Minors. Graph G is called a minor of graph H is if G can be obtained from H by a series of edge deletions and edge contractions. Graph H is also called a G-minor Example: K 5 is a minor of the Petersen graph P, but P does not contain a K 5 -subdivision. 6

9 The Graph Minor Theorem Theorem. (Robertson and Seymour, ) In any infinite list of graphs, some graph is a minor of another. Proof: more than 500 pages in 20 papers. Corollary For any graph property that is closed under taking minors, there exists finitely many minimal forbidden minors. Homework. Wagner s Theorem. Every nonplanar graph contains either a K 5 or K 3,3 -minor. For embeddability on the projective plane, it is known that there are 35 minimal forbidden minors. For embeddability on the torus, we don t know the exact number of minimal forbidden minors; there are more than 800 known. 7

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