INF562, Lecture 4: Geometric and combinatorial properties of planar graphs

Transcription

1 INF562, Lecture 4: Geometric and combinatorial properties of planar graphs 5 jan 2016 Luca Castelli Aleardi

2 Intro Graph drawing: motivations and applications

3 Graph drawing and data visualization Global transportation system

4 Graph drawing and data visualization Roads, railways,...

5 Graph drawing and data visualization Social network graph Community detection (via graph clustering)

6 Design of integrated circuits (VLSI) Planar graphs french roads network

7 Design of integrated circuits (VLSI) Planar graphs french roads network 9 accidents en 2012 (last one, on 28th sptember)

8 Meshes and graphs in computational geometry Delaunay triangulations, Voronoi diagrams, planar meshes,... triangles meshes already used in early 19th century (Delambre et Mchain) GIS Technology Delaunay triangulation Planar mesh by L. Rineau, M. Yvinec Terrain modelling Spherical Parameterization (Sheffer Gotsman) Voronoi diagram

9 Mesh parameterization (and straight line drawing) given a (planar) graph G compute a mapping ρ : (V G ) R 2 s.t. edges are straight line segments without crossings (given a 3D mesh M compute a bijective mapping ρ : (V G ) R 2 ) ρ(v i ) = (x i, y i )

10 Mesh parameterization: applications Texture mapping and morphing (pictures by A. Sheffer)

11 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G =

12 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G = adjacency matrix A G [i, j] = { 0 1 if v i is adjacent to v j otherwise

13 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G =

14 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G =

15 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G =

16 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G =

17 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G =

18 Graph drawing: motivation Challenge: what kind of graph does A G represent? A G =

19 Part I Major results in graph theory

20 Major results (on planar graphs) in graph theory

21 Major results (on planar graphs) in graph theory Kuratowski theorem (1930) (cfr Wagner s theorem, 1937) G contains neither K 5 nor K 3,3 as minors

22 Major results (on planar graphs) in graph theory Kuratowski theorem (1930) (cfr Wagner s theorem, 1937) G contains neither K 5 nor K 3,3 as minors Fáry theorem (1947) v 2 Every (simple) planar graph admits a straight line planar embedding (no edge crossings) v 1 v 4 v 2 v 1 v 4 v 1 v 3 v 5 dessin non planaire de G v v 3 v 2 5 un dessin planaire de G v 3

23 Major results (on planar graphs) in graph theory Thm (Steinitz, 1916) Fáry theorem (1947) v 2 Every (simple) planar graph admits a straight line planar embedding (no edge crossings) v 1 v 4 v 2 v 1 v 4 v 1 v 3 v 5 dessin non planaire de G v v 3 v 2 5 un dessin planaire de G v 3

24 Major results (on planar graphs) in graph theory Thm (Steinitz, 1916) 3-connected planar graphs are the 1- skeletons of convex polyhedra Thm (Whitney, 1933) 3-connected planar graphs admit a unique planar embedding (up to homeomorphism and inversion of the sphere).

25 Major results (on planar graphs) in graph theory Thm (Steinitz, 1916) 3-connected planar graphs are the 1- skeletons of convex polyhedra Def G is 3-connected if is connected and the removal of one or two vertices does not disconnect G Thm (Whitney, 1933) 3-connected planar graphs admit a unique planar embedding (up to homeomorphism and inversion of the sphere). at least 3 vertices are required to disconnect the graph

26 Major results (on planar graphs) in graph theory Thm (Koebe-Andreev-Thurston) Every planar graph with n vertices is isomorphic to the intersection graph of n disks in the plane. Schnyder woods (via dimension of partial orders) dim(g) 3 Thm (Colin de Verdière, 1990) Theorem (Lovasz Schrijver 99) Colin de Verdiere invariant (multiplicity of λ 2 eigenvalue of a generalized laplacian) µ(g) deg(v i ) L G [i, k] = { AG [i, j] Given a 3-connected planar graph G, the eigenvectors ξ 2, ξ 3, ξ 4 of a CdV matrix defines a convex polyhedron containing the origin.. M v N S 2 v N

27 Major results (on planar graphs) in graph theory ρ : (V G ) R 2 is barycentric iff for each inner node v i, ρ(v i ) is the barycenter of the images of its neighbors N(v 4 ) = {v 1, v 2, v 3, v 5 } N(v 5 ) = {v 2, v 3, v 4 } v 2 v 3 v 1 v 4 v 5 Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a barycentric representation ρ in R 2. v 2 v 1 v 3 ρ(v i ) = j N(i) w ij ρ(v j ) ( j w ij = 1 and w ij > 0 Get a straight line drawing solving a system a linear equations L = laplacian matrix { M x = a x M y = a y

28 Major results (on planar graphs) in graph theory Theorem (Schnyder 89) A graph G is planar if and only if the dimension of its incidence poset is at most 3 Theorem (Schnyder, Soda 90) For a triangulation T having n vertices, we can draw it on a grid of size (2n 5) (2n 5), by setting v 0 = (2n 5, 0), v 1 = (0, 0) and v 2 = (0, 2n 5). (a) v 2 (b) v 5 v 0 v 1 v 4 v 3 v 0 v 1 v 3 v 4 v 5 (1, 2, 4) (2, 4, 1) (4, 1, 2)

29 Part II What is a surface mesh? (a short digression on embedded graphs, simplicial complexes and topological and combinatorial maps)

30 What is a (surface) mesh? surface mesh: set of vertices, edges and faces (polygons) defining a polyhedral surface in embedded in 3D (discrete approximation of a shape) Combinatorial structure + geometric embedding Connectivity : the underlying map incidence relations between triangles, vertices and edges vertex coordinates

31 Planar and surface meshes: definition planar triangulation embedded in R 3 triangle mesh planar triangulation spherical drawing planar map straight line drawing of a dodecahedron spherical parameterizations of a triangle mesh (Gotsman, Gu Sheffer, 2003) toroidal map (Eric Colin de Verdière)

32 Surface meshes as simplicial complexes abstract simplicial complex K (set of simplices) V = {v 0, v 1,..., v n 1 } E = {{i, j}, {k, l},...} F = {{i, j, k}, {j, i, l},...} inclusion property: ρ K and σ ρ σ K not valid simplicial complex intersection property: given two simplices σ 1, σ 2 of K, the intersection σ 1 σ 2 is a face of both

33 A graph G = (V, E) is a pair of: Surface meshes as (topological) maps (geomettric realizations of maps) a set of vertices V = (v 1,..., v n ) two different embeddings of the same graph a collection of E = (e 1,..., e m ) elements of the cartesian product V V = {(u, v) u V, v V } (edges). a planar drawing is a cellular embedding of G into R 2, satisfying: cellular embeddings of a graph defining the same (planar) map (i) graph vertices are represented as points ; (ii) edges are represented as no croissing curves ; (iii) faces are simply connected. (topologica) map: cellular embedding up to homeomorphism (equivalence class) two cellular embeddings defining the same planar map

34 Surface meshes as combinatorial maps (geomettric realizations of maps) permutations on the set H of the 2n half-edges (i) α involution without fixed point; (ii) ασφ = Id; (iii) the group generated by σ, α et φ transitively on H. φ = (1, 2, 3, 4)(17, 23, 18, 22)(5, 10, 8, 12)(21, 19, 24, 15)... α = (2, 18)(4, 7)(12, 13)(9, 15)(14, 16)(10, 23)...

35 Surface meshes as combinatorial maps (geomettric realizations of maps) permutations on the set H of the 2n half-edges (i) α involution without fixed point; (ii) ασφ = Id; (iii) the group generated by σ, α et φ transitively on H. φ = (1, 2, 3, 4)(17, 23, 18, 22)(5, 10, 8, 12)(21, 19, 24, 15)... α = (2, 18)(4, 7)(12, 13)(9, 15)(14, 16)(10, 23)...

36 Half-edge data structure: polygonal (orientable) meshes e f + 5 h + n 2n + 5 (2e) + n = 32n + n Size (number of references) 1 class Point{ double x; double y; } geometric information vertex(e) next(e) opposite(e) e face(e) prev(e) class Halfedge{ Halfedge prev, next, opposite; Vertex v; Face f; } class Vertex{ Halfedge e; Point p; } class Face{ Halfedge e; } combinatorial information

37 Mesh representations: classification Manifold meshes non manifold or non orientable meshes triangle meshes no boundary with boundaries genus 1 mesh Every edge is shared by at most 2 faces For every vertex v, the incident faces form an open or closed fan quad meshes polygonal meshes v

38 Part III Euler formula and its consequences

39 Euler-Poincaré characteristic: topological invariant χ := n e + f planar map n e + f = 2 Euler s relation

40 Euler-Poincaré characteristic: topological invariant χ := n e + f n e + f = 2 Euler s relation planar map (convex) polyhedron χ = 0 χ = 4 n = 1660 e = 4992 f = 3328 g = 3 n e + f = 2 2g n = 364 e = 675 f = 302 b = 11 g = 0 n e + f = 2 b

41 Euler s relation for polyhedral surfaces polyedre, n sommets Overview of the proof n e + f = 2 carte planaire arbre couvrant n 1 aretes patron f 1 aretes arbre couvrant du dual

42 Euler s relation for polyhedral surfaces Overview of the proof n e + f = 2 polyedre, n sommets f faces carte planaire arbre couvrant n 1 aretes patron e = (n 1) + (f 1) f 1 aretes f sommets arbre couvrant du dual

43 Using spherical geometry An alternative geometric proof of Euler Formula n e + f = 2 q γ C p O A α β B γ δ α β Geodesic: arc of great circle (between p and q) Girard Theorem α + β + γ = π + area(a,b,c) r 2 i k α i = (k i 2)π + area(p i) r 2 Spherical drawing of an octahedron α+β+γ+δ = 2π Theorem (Girard) Given a spherical polygon P i of size k i, the sum of internal angles α i satisfy

44 Using spherical geometry An alternative geometric proof of Euler Formula n e + f = 2 q γ C p O A α β B γ δ α β Geodesic: arc of great circle (between p and q) Girard Theorem α + β + γ = π + area(a,b,c) r 2 i k α i = (k i 2)π + area(p i) r 2 Spherical drawing of an octahedron α+β+γ+δ = 2π Theorem (Girard) Given a spherical polygon P i of size k i, the sum of internal angles α i satisfy d j=1 α j = 2π for each v i area(p i) = 4πr 2

45 Euler s relation for polyhedral surfaces Corollary: linear dependence between edges, vertices and faces f 2n 4 proof (double counting argument) e 3n 6 f = f 1 + f 2 + f n = n 1 + n 2 + n all faces have degree at least 3 (G simple simple), then we get f = f 3 + f every edge appears twice 2e = 3 f f then we get 2e 3f 0

46 Euler s relation for polyhedral surfaces Corollary: linear dependence between edges, vertices and faces f 2n 4 e 3n 6 given 2e 3f 0 by applying Euler formula, we obtain 3n 6 = 3(e f + 2) = 3e 3f 0

47 Major results: Kuratowski theorem Kuratowski theorem (1930) (cfr Wagner s theorem, 1937) G is planar iff it does not contain K 5 nor K 3,3 as minors proof (one direction) K 3,3 bipartite: no cycle of length 3 e 2n 4 = 8 < 9 e 3n 6 = 9 but we have e(k 5 ) = ( 5 2) = 10

48 Part IV Tutte s planar embedding

49 Preliminaries: barycentric coordinates q = n i α i v i (avec i α i = 1) coefficients (α 1,..., α n ) are called barycentric coordinates of q (relative to v 1,..., v n ) Geometric interpretation of barycentric coordinates w = 0 v 2 α 1 α 0 q α 2 v 0 v 1 q = α 0 v 0 + α 1 v 1 + α 2 v 2 q = area(v,v 1,v 2 )v 0 +area(v 0,v,v 2 )v 1 +area(v 0,v 1,v)v 2 area(v 0,v 1,v 2 ) v 2 q v 0 v 1 u < 0, v > 0, w > 0 (u, v, w) = (0, 1, 0) (0, 0, 1) (1, 0, 0) v = 0 u > 0, v < 0, w > 0 u = 0

50 Tutte s theorem Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a convex representation ρ in R 2.

51 Tutte s theorem Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a convex representation ρ in R 2. ρ : (V G ) R 2 ρ is convex the images of the faces of G are convex polygons

52 Tutte s theorem Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a convex representation ρ in R 2. ρ : (V G ) R 2 ρ is barycentric the images of interior vertices are barycenters of their neighbors ρ(v i ) = N(v 4 ) = {v 1, v 2, v 3, v 5 } j N(i) w ij ρ(v j ) where w ij satisfy j w ij = 1, and w ij > 0 according to Tutte: w ij = 1 deg(v i )

53 Tutte s spring embedder (physical interpretation) ρ minimizes Lemma The representation ρ with minimum energy is uniquely determined (in R d ).: we require G connected and F not empty ρ : (V G ) R 2 E(ρ) := ρ(v i ) ρ(v j ) 2 = { (i,j) E (i,j) E E(ρ) subject to ρ(v k ) = p k = (x k, y k ) (for exterior vertices v k ) The energy E(ρ) is strictly convex (x i x j ) 2 + (y i y j ) 2 (ρ(v i ) ρ(v j )) = 0 j N(i) ρ(v i ) = 1 d i j N(i) ρ(v j )

54 Tutte s theorem: main steps chose a cycle F (the outer face of G) in the right way a cycle such that G \ F is connected (deletion of vertices and edges)

55 Tutte s theorem: main steps chose a cycle F (the outer face of G) in the right way a cycle such that G \ F is connected (deletion of vertices and edges) choose a convex polygon P of size k = F such that ρ(f ) = P v 2 v 1 v 4 v 3 v 5

56 Tutte s theorem: main steps chose a cycle F (the outer face of G) in the right way a cycle such that G \ F is connected (deletion of vertices and edges) choose a convex polygon P of size k = F such that ρ(f ) = P v 2 v 1 v 4 solve equations for images of inner vertices ρ(v i ): v 3 v 5 ρ(v i ) = w ij ρ(v j ) j N(i) ρ(v i ) w ij ρ(v j ) = 0 j N(i) according to Tutte: w ij = 1 deg(v i )

57 Tutte s theorem: main steps chose a cycle F (the outer face of G) in the right way a cycle such that G \ F is connected (deletion of vertices and edges) choose a convex polygon P of size k = F such that ρ(f ) = P v 2 v 1 v 4 solve two linear systems: { (I W ) x = b x (I W ) y = b y { v 3 v 5 ρ x (v i ) w ij ρ x (v j ) = 0 j N(i) ρ y (v i ) w ij ρ y (v j ) = 0 j N(i)

58 Tutte s theorem: main steps chose a cycle F (the outer face of G) in the right way a cycle such that G \ F is connected (deletion of vertices and edges) choose a convex polygon P of size k = F such that ρ(f ) = P v 2 v 1 v 4 v 3 v 5 solve a linear system:

59 Validity of Tutte s theorem: main results show that the linear system admit a (unique) solution: { (I W ) x = b x (I W ) y = b y matrix (I W ) is inversible a barycentric drawing is planar: no edge crossing a 3-connected planar graph G has a peripheral cycle Claim (existence of peripheral cycles) In a 3-connected planar graph peripheral cycles are exactly the faces (of the embedding) v 2 v 3 v 1 v 4 v 5

60 Advantages of Tutte s drawing the drawing is guaranteed to be planar (no edge crossing) no need of the map structure graph structure + a peripheral cycle v 2 v 3 v 1 v 4 v 5 very easy to implement: no need of sophisticated data structure or preprocessing linear systems to solves nice drawings (detection of symmetries)

61 Drawbacks of Tutte s drawing requires to solve linear systems of equations (of size n) { (I W ) x = b x (I W ) y = b y complexity O(n 3 ) or O(n 3/2 ) with methods more involved exponential size of the resulting vertex coordinates (with respect to n) drawings are not always nice

62 Tutte s spring embedder: iterative version choose an outer face F, and a convex polygon P put exterior vertices v F on the polygon repeat (until convergence) for each inner vertex v V i compute x v = 1 deg(v) y v = 1 deg(v) (u,v) E x u (u,v) E y u V i inner vertices (u, v) edge connecting v and u

63 Graph drawing: force directed paradigm points are randmoly distributed in the plane repeat (until convergence) for each vertex v V compute v = v + c 4 F(v) où F(v) := F a (v) + F r (v) attractive force (between adjacent vertices) F a (v) = c 1 (u,v) E log(d uv)/c 2 ) repulsive force (between non adjacent vertices) F r (v) = c 3 u V 1 d 2 uv c 1 = 2 c 2 = 1 c 3 = 1 c 4 = 0.1

64 Part V Tutte s theorem: the proof First: existence and uniqueness of barycentric representations Second: the barycentric representation defines a planar drawing (no edge crossing) Third: existence and computation of peripheral cycles

65 (Some notions of) Spectral graph theory Laplacian matrix (simple graphs) Q G [i, k] = { deg(v i ) AG [i, j] if i = j otherwise G v 2 v 3 v 1 v 5 v 4 Q G =

66 (Some notions of) Spectral graph theory Laplacian matrix (counting multiple edges) G v 1 v 2 Q G { deg(v i ) [i, k] = edges from vi to v j if i = j otherwise v 3 v 5 v 4 Q G [i 1, i 2,...] = Q G \ { line i 1, line i 2, column i 1, column i 2, G v 1 v 4 Q G = Q G = v 5

67 (Some notions of) Spectral graph theory Lemma (Laplacian and the number of spanning trees) Let Q be the laplacian of a graph G, with n vertices. Then the number of spannig trees of G is: τ(g) = det(q[i]) (i n) v 1 G v 5 v 4 Q G Q G [1] = = 11

68 First: existence and uniqueness of barycentric representations

69 First: existence and uniqueness of barycentric representations Theorem Let G be a 3-connected planar graph with n vertices, and F a peripheral cycle (such that G \ F is connected). Let P be a convex polygon, such that ρ(f ) = P. Then the barycentric representation ρ exists (and is unique) Goal: show that the two systems above admit a solution (unique) { (I W ) x = b x (I W ) y = b y x = [x 1, x 2,..., x n k ] y = [y 1, y 2,..., y n k ] (coordinates of inner vertices) ρ(v i ) = ρ(v i ) = n w ij ρ(v j ) i = 1,..., (n k) 1 j N(i) w ij ρ(v j ) (one equation for each inner vertex)

70 First: existence and uniqueness of barycentric representations Proof ρ(v i ) j N(i) w ij ρ(v j ) = 0 { (I W ) x = b x (I W ) y = b y

71 First: existence and uniqueness of barycentric representations Proof ρ(v i ) j N(i) w ij ρ(v j ) = 0 { (I W ) x = b x (I W ) y = b y x 4 x 5 = b 4x b 5x v 2 v 1 v 4 v 1 N(v 4 ) = {v 1, v 2, v 3, v 5 } ρ(v i ) := (x i, y i ) y 4 y 5 = b 4y b 5y v 3 v 5 v 2 v 3 N(v 5 ) = {v 2, v 3, v 4 }

72 First: existence and uniqueness of barycentric representations Proof ρ(v i ) j N(i) w ij ρ(v j ) = 0 deg(v i )ρ(v i ) j N(i) ρ(v j ) = 0 { (I W ) x = b x (I W ) y = b y { M x = a x M y = a y x 4 x 5 = b 4x b 5x y 4 y 5 = b 4y b 5y M = { ρ(v 4 ) 1 4 ρ(v 5) = 1 4 ρ(v 1) ρ(v 2) ρ(v 3) 1 3 ρ(v 4) + ρ(v 5 ) = 1 3 ρ(v 2) ρ(v 3)

73 First: existence and uniqueness of barycentric representations Q G = Q G [1, 2, 3] := M = { ρ(v 4 ) 1 4 ρ(v 5) = 1 4 ρ(v 1) ρ(v 2) ρ(v 3) 1 3 ρ(v 4) + ρ(v 5 ) = 1 3 ρ(v 2) ρ(v 3)

74 First: existence and uniqueness of barycentric representations deg(v i )ρ(v i ) j N(i) ρ(v j ) = 0 G G/F det(m) = τ(q G/F ) > 0 G/F is connected { M x = a x M y = a y v 2 v 1 G/F = G/{v 1, v 2, v 3 } v 123 v 4 v 4 Q G = v 3 v 5 Q G [1, 2, 3] := M = { ρ(v 4 ) 1 4 ρ(v 5) = 1 4 ρ(v 1) ρ(v 2) ρ(v 3) 1 3 ρ(v 4) + ρ(v 5 ) = 1 3 ρ(v 2) ρ(v 3) v 5 Q G/F

75 First: existence and uniqueness of barycentric representations deg(v i )ρ(v i ) j N(i) ρ(v j ) = 0 M x = a x M y = a y G G/F det(m) = τ(q G/F ) > 0 G/F is connected M admits inverse { v 2 v 1 G/F = G/{v 1, v 2, v 3 } v 123 v 4 v 4 Q G = v 3 v 5 Q G [1, 2, 3] := M = { ρ(v 4 ) 1 4 ρ(v 5) = 1 4 ρ(v 1) ρ(v 2) ρ(v 3) 1 3 ρ(v 4) + ρ(v 5 ) = 1 3 ρ(v 2) ρ(v 3) v 5 Q G/F

76 Second: the barycentric representation defines a planar drawing Theorem Let G be a 3-connected planar graph with n vertices, and F a peripheral cycle (such that G \ F is connected). Let P be a convex polygon, such that ρ(f ) = P. Then the barycentric representation defines a planar drawing (no edge crossing)

77 Second: the barycentric representation defines a planar drawing Theorem Let G be a 3-connected planar graph with n vertices, and F a peripheral cycle (such that G \ F is connected). Let P be a convex polygon, such that ρ(f ) = P. Then the barycentric representation defines a planar drawing (no edge crossing) ρ(v 3 ) ρ(v 5 ) ρ(v 2 ) ρ(v 4 ) ρ(v) ρ(v 8 ) ρ(v 1 ) ρ(v 7 ) p 3 p 5 p 4 p 2 p p 8 p 7 p 1 ρ(v 6 ) p 6 barycentric representation of G planar drawing G

78 Second: the barycentric representation defines a planar drawing Theorem Let G be a 3-connected planar graph with n vertices, and F a peripheral cycle (such that G \ F is connected). Let P be a convex polygon, such that ρ(f ) = P. Then the barycentric representation defines a planar drawing (no edge crossing) ρ(v 3 ) ρ(v 5 ) ρ(v 2 ) ρ(v 4 ) ρ(v) ρ(v 6 ) ρ(v 8 ) ρ(v 1 ) ρ(v 7 ) σ(v) α(v) dessin non planaire de G σ(v) = α(v) p 3 p 5 p 4 p 2 p p 6 p 8 p 7 planar drawing of G α(v) = 2π p 1

79 Second: the barycentric representation defines a planar drawing Claim 1 σ(v) 2π = α(v) ρ(v 3 ) ρ(v 5 ) ρ(v 2 ) ρ(v 4 ) ρ(v) ρ(v 6 ) ρ(v 8 ) ρ(v 1 ) ρ(v 7 ) σ(v) := k γ k k γ i 1 γ j γ i γ i+1 i j k β γ δ j i β + γ + δ = 2π

80 Second: the barycentric representation defines a planar drawing Claim 2 v α(v) v σ(v) = πf k p 3 p 5 p 4 α β γ p 2 δ p 6 η p 1 p 8 p 7 α + β + γ + δ +... = ( F 2)π β + γ + δ = π i β γ δ j v α(v) = α(v) + inner vertices outer vertices α(v) = 2π V \ F + ( F 2)π v σ(v) = πf v V \ F v F sum over inner and outer vertices sum of the angles of triangles (3 angles per face)

81 Second: the barycentric representation defines a planar drawing Conclusion Euler formula α(v) = σ(v) n (e + F ) + f = ( V \ F + F ) (e + F ) + (t + 1) Claim 2 inner edges Counts the number of edges Claim 1 } inner triangles 3t = 2e + F } v v α(v) := 2π V \ F + ( F 2)π = πf v α(v) v σ(v) = πf v α(v) = v σ(v) 2π = α(v) σ(v) } 2π V \ F + ( F 2)π = πf α(v) = v σ(v) α(v) = σ(v) = 2π

82 Third: peripheral cycles (non separating cycles) Definition A non-separating cycle C of a graph G is an induced cycle in G such that G \ C is connected (after the deletion of edges and vertices of C) peripheral cycle v 2 v 3 v 1 v 4 v 5 separating cycle v 1 v 4 A peripheral cycle C is a simple cycle of a connected graph G such that for any two edges e 1 and e 2 in G \ C there exists a path (starting at e 1 and ending at e 2 ) with no interior vertices on C v 2 v 3 v 5 no peripheral cycles K 2,3

83 Third: existence of peripheral cycles Lemma In a 3-connected planar graph peripheral cycles are exactly the faces (of the embedding) proof (exercise)