INF562, Lecture 4: Geometric and combinatorial properties of planar graphs
5 jan 2016
Luca Castelli Aleardi
1 Intro Graph drawing: motivations and applications
2 Graph drawing and data visualization Global transportation system
3 Graph drawing and data visualization Roads, railways, ...
4 Graph drawing and data visualization Social network graph
Community detection (via graph clustering)
5 Planar graphs french roads network Design of integrated circuits (VLSI)
6 Planar graphs french roads network Design of integrated circuits (VLSI)
www.2m40.com
9 accidents en 2012 (last one, on 28th sptember)
6 Meshes and graphs in computational geometry Delaunay triangulations, Voronoi diagrams, planar meshes, ...
GIS Technology triangles meshes already used in early 19th century (Delambre et Mchain)
Delaunay triangulation Terrain modelling Planar mesh by L. Rineau, M. Yvinec
Spherical Parameterization (Sheffer Gotsman)
Voronoi diagram
7 Mesh parameterization (and straight line drawing) 2 given a (planar) graph G compute a mapping ρ :(VG) −→ R s.t. edges are straight line segments without crossings
2 (given a 3D mesh M compute a bijective mapping ρ :(VG) −→ R )
ρ(vi) = (xi, yi)
8 Mesh parameterization: applications Texture mapping and morphing
(pictures by A. Sheffer)
9 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
10 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
adjacency matrix
1 if vi is adjacent to vj AG[i, j] = { 0 otherwise
10 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
10 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
10 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
10 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
10 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
10 Graph drawing: motivation
0 1 1 0 1 1 Challenge: what kind of graph does AG represent?
1 0 1 1 0 1 1 1 0 1 1 0 AG = 0 1 1 0 1 1 1 0 1 1 0 1
1 1 0 1 1 0
10 Part I Major results in graph theory
11 Major results (on planar graphs) in graph theory
12 Major results (on planar graphs) in graph theory
Kuratowski theorem (1930) (cfr Wagner’s theorem, 1937)
• G contains neither K5 nor K3,3 as minors
12 Major results (on planar graphs) in graph theory
Kuratowski theorem (1930) (cfr Wagner’s theorem, 1937)
• G contains neither K5 nor K3,3 as minors
F´arytheorem (1947) • Every (simple) planar graph admits a straight line planar embedding (no edge crossings) v1 v1 v1 v2 v2 v4 v4
v2 v3 v5 v3 v5 dessin non planaire de G un dessin planaire de G v3
12 Major results (on planar graphs) in graph theory Thm (Steinitz, 1916)
F´arytheorem (1947) • Every (simple) planar graph admits a straight line planar embedding (no edge crossings) v1 v1 v1 v2 v2 v4 v4
v2 v3 v5 v3 v5 dessin non planaire de G un dessin planaire de G v3
12 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).
12 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 Def G is 3-connected if planar embedding (up to homeomorphism and is connected and inversion of the sphere). the removal of one or two vertices does not disconnect G
at least 3 vertices are required to disconnect the graph
12 Major results (on planar graphs) in graph theory Thm (Koebe-Andreev-Thurston) Every planar graph with n vertices is isomorphic to Schnyder woods (via the intersection graph of n disks in the plane. dimension of partial or- ders) • dim(G) ≤ 3
Thm (Colin de Verdi`ere,1990) Theorem (Lovasz Schrijver ’99) Given a 3-connected planar graph G, the eigenvectors ξ , ξ , ξ of a CdV Colin de Verdiere invariant (multiplicity of λ eigen- 2 3 4 2 matrix defines a convex polyhedron containing the origin.. value of a generalized laplacian) N • µ(G) ≤ 3 N S2 v M v 4 −1 ...... 0 deg(vi) −1 5 ... LG[i, k] = ...... { −AG[i, j] ...... 0 ... 3
12 Major results (on planar graphs) in graph theory Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a barycentric representation ρ in R2.
X P ρ(vi) = wijρ(vj) ( j wij = 1 and wij > 0 j∈N(i) 2 ρ :(VG) −→ R is barycentric iff for each inner node vi, ρ(vi) is the barycenter of Get a straight line drawing the images of its neighbors solving a system a linear equations
N(v4) = {v1, v2, v3, v5}
N(v5) = {v2, v3, v4} 3 −1 −1 −1 0 −1 4 −1 −1 −1 M · x = ax v1 v1 L = −1 −1 4 −1 −1 v2 { M · y = ay v4 −1 −1 −1 4 −1 0 −1 −1 −1 3 v2 v3 v5 laplacian matrix v3
12 Major results (on planar graphs) in graph theory Theorem (Schnyder ’89) A graph G is planar if and only if the di- mension 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 v0 = (2n − 5, 0), v1 = (0, 0) and v2 = (0, 2n − 5).
v2
(a) (b) v5
v4
v3 v0 v1 v3 (1, 2, 4)
v4 (2, 4, 1) v0 v1 v5 (4, 1, 2)
12 Part II What is a surface mesh?
(a short digression on embedded graphs, simplicial complexes and topological and combinatorial maps)
13 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
vertex coordinates incidence relations between triangles, vertices and edges
14 Planar and surface meshes: definition
planar triangulation planar triangulation planar map embedded in R3 spherical drawing straight line drawing of a dodecahedron triangle mesh
spherical parameterizations of a triangle mesh (Gotsman, Gu Sheffer, 2003) toroidal map (Eric Colin de Verdiere)`
15 Surface meshes as simplicial complexes abstract simplicial complex K (set of simplices)
V = {v0, v1, . . . , vn−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 σ1, σ2 of K, the intersection σ1 ∪ σ2 is a face of both
16 Surface meshes as (topological) maps (geomettric realizations of maps)
two different embeddings of the same graph A graph G = (V,E) is a pair of:
• a set of vertices V = (v1, . . . , vn)
• a collection of E = (e1, . . . , em) elements of the cartesian product V × V = {(u, v) | u ∈ V, v ∈ V } (edges). cellular embeddings of a graph defining the same (planar) map a planar drawing is a cellular embedding of G into R2, satisfying: (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
17 Surface meshes as combinatorial maps (geomettric realizations of maps) 17 22 23 18 3 permutations on the set H of the 2n half-edges 20 2 10 19 11 6 1 3 5 8 21 24 (i) α involution without fixed point; 14 4 12 15 (ii) ασφ = Id; 7 16 13 9 (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) ...
18 Surface meshes as combinatorial maps (geomettric realizations of maps) 17 22 23 18 3 permutations on the set H of the 2n half-edges 20 2 10 19 11 6 1 3 5 8 21 24 (i) α involution without fixed point; 14 4 12 15 (ii) ασφ = Id; 7 16 13 9 (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) ...
18 Half-edge data structure: polygonal (orientable) meshes
8 5 6 9
3 7 0 e
4 11 2 10
1 class Point{ double x; double y; f + 5 × h + n ≈ 2n + 5 × (2e) + n = 32n + n }
Size (number of references) geometric information
vertex(e) opposite(e) class Halfedge{ Halfedge prev, next, opposite; Vertex v; e Face f; next(e) } class Vertex{ prev( ) Halfedge e; face(e) e Point p; } class Face{ Halfedge e; } combinatorial information
19 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 inci- dent faces form an open or closed fan quad meshes polygonal meshes v
20 Part III Euler formula and its consequences
21 Euler-Poincare´ characteristic: topological invariant χ := n − e + f
n − e + f = 2 Euler’s relation planar map
22 Euler-Poincare´ characteristic: topological invariant χ := n − e + f
n − e + f = 2 Euler’s relation planar map (convex) polyhedron
n = 1660 χ = 0 χ = −4 e = 4992 f = 3328 n − e + f = 2 − 2g g = 3
n = 364 e = 675 f = 302 n − e + f = 2 − b b = 11 g = 0
23 Euler’s relation for polyhedral surfaces Overview of the proof n − e + f = 2
polyedre, n sommets
carte planaire arbre couvrant patron n − 1 aretes
f − 1 aretes
arbre couvrant du dual
24 Euler’s relation for polyhedral surfaces Overview of the proof n − e + f = 2
polyedre, n sommets f faces
carte planaire arbre couvrant patron n − 1 aretes
f − 1 aretes f sommets e = (n − 1) + (f − 1) arbre couvrant du dual
25 Using spherical geometry n − e + f = 2 An alternative geometric proof of Euler Formula
C q γ
γ p β δ α O A α β B
Spherical drawing of an octahe- Geodesic: arc of great circle dron Girard Theorem α+β+γ+δ = 2π (between p and q) area(A,B,C) α + β + γ = π + r2 Theorem (Girard) Given a spherical polygon Pi of size ki, the sum of internal angles αi satisfy
P area(Pi) i≤k αi = (ki − 2)π + r2
26 Using spherical geometry n − e + f = 2 An alternative geometric proof of Euler Formula
C q γ
γ p β δ α O A α β B
Spherical drawing of an octahe- Geodesic: arc of great circle dron Girard Theorem α+β+γ+δ = 2π (between p and q) area(A,B,C) α + β + γ = π + r2 Theorem (Girard) Given a spherical polygon Pi of size ki, the sum of internal angles αi satisfy
P area(Pi) i≤k αi = (ki − 2)π + r2 Pd j=1 αj = 2π for each v P 2 i area(Pi) = 4πr
27 Euler’s relation for polyhedral surfaces Corollary: linear dependence between edges, vertices and faces f ≤ 2n − 4 e ≤ 3n − 6 proof (double counting argument)
f = f1 + f2 + f3 + ... n = n1 + n2 + n3 + ... all faces have degree at least 3 (G simple simple), then we get f = f3 + f4 + ... every edge appears twice 2e = 3 · f3 + 4 · f4 + ... then we get 2e − 3f ≥ 0
28 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
29 Major results: Kuratowski theorem Kuratowski theorem (1930) (cfr Wagner’s theorem, 1937) G is planar iff it does not contain K5 nor K3,3 as minors
proof (one direction)
K3,3 bipartite:
no cycle of length 3 e ≤ 3n − 6 = 9
e ≤ 2n − 4 = 8 < 9 5 but we have e(K5) = 2 = 10
30 Part IV Tutte’s planar embedding
31 Preliminaries: barycentric coordinates Pn P q = i αivi (avec i αi = 1)
coefficients (α1, . . . , αn) are called barycentric coordinates of q (relative to v1, . . . , vn) Geometric interpretation w = 0 of barycentric coordinates v2 v2 (u, v, w) = (0, 1, 0) q u < 0, v > 0, w > 0 α1 α0 q
α2 v0 v1 (0, 0, 1) (1, 0, 0) v = 0 v0 v1
q = α0v0 + α1v1 + α2v2 u > 0, v < 0, w > 0 q = area(v,v1,v2)v0+area(v0,v,v2)v1+area(v0,v1,v)v2 u = 0 area(v0,v1,v2)
32 Tutte’s theorem Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a convex representation ρ in R2.
33 Tutte’s theorem Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a convex representation ρ in R2.
2 ρ :(VG) −→ R
ρ is convex the images of the faces of G are convex polygons
33 Tutte’s theorem Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a convex representation ρ in R2. 2 ρ :(VG) −→ R ρ is barycentric the images of interior vertices are barycenters of their neighbors
P X where wij satisfy j wij = 1, and wij > 0 ρ(vi) = wijρ(vj) 1 according to Tutte: wij = j∈N(i) deg(vi)
N(v4) = {v1, v2, v3, v5}
33 Tutte’s spring embedder (physical interpretation) The representation ρ with minimum energy is uniquely determined (in Rd).: we require G connected and F not empty 2 ρ :(VG) −→ R
X 2 X 2 2 E(ρ) := |ρ(vi) − ρ(vj)| = (xi − xj) + (yi − yj) (i,j)∈E (i,j)∈E ρ minimizes E(ρ)
{ subject to ρ(vk) = pk = (xk, yk) (for exterior vertices vk) Lemma E(ρ) X The energy is strictly convex (ρ(vi) − ρ(vj)) = 0 j∈N(i)
1 X ρ(vi) = ρ(vj) di j∈N(i)
34 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)
35 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)
v1 choose a convex polygon P of size k = |F | v2 such that ρ(F ) = P v4
v3 v5
35 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)
v1 choose a convex polygon P of size k = |F | v2 such that ρ(F ) = P v4
v3 v5 solve equations for images of inner vertices ρ(vi):
X ρ(vi) = wijρ(vj) j∈N(i) X ρ(vi) − wijρ(vj) = 0 1 according to Tutte: wij = j∈N(i) deg(vi)
35 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)
v1 choose a convex polygon P of size k = |F | v2 such that ρ(F ) = P v4
v3 v5
solve two linear systems: X ρx(vi) − wijρx(vj) = 0 (I − W ) · x = bx j∈N(i) { (I − W ) · y = by X { ρy(vi) − wijρy(vj) = 0 j∈N(i)
35 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)
v1 choose a convex polygon P of size k = |F | v2 such that ρ(F ) = P v4
v3 v5
solve a linear system:
35 Validity of Tutte’s theorem: main results show that the linear system admit a (unique) solution: (I − W ) · x = b x matrix (I − W ) is inversible { (I − W ) · y = by
a barycentric drawing is planar: no edge crossing
a 3-connected planar graph G has a peripheral cycle v1 Claim (existence of peripheral cycles) v2 v4 In a 3-connected planar graph peripheral cycles are exactly the faces (of the embedding) v3 v5
36 Advantages of Tutte’s drawing the drawing is guaranteed to be planar (no edge crossing) v1 v2 v4 no need of the map structure v v5 graph structure + a peripheral cycle 3
very easy to implement: no need of sophisticated data structure or preprocessing
linear systems to solves
nice drawings (detection of symmetries)
37 Drawbacks of Tutte’s drawing
requires to solve linear systems of equations (of size n) (I − W ) · x = bx complexity O(n3) 3/2 { (I − W ) · y = by or O(n ) with methods more involved
exponential size of the resulting vertex coordinates (with respect to n)
drawings are not always ”nice”
38 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 ∈ Vi compute 1 P Vi inner vertices xv = (u,v)∈E xu deg(v) (u, v) edge connecting v and u 1 P yv = deg(v) (u,v)∈E yu
39 Graph drawing: force directed paradigm
points are randmoly distributed in the plane repeat (until convergence) for each vertex v ∈ V compute
v = v + c4 · F(v)
ou` F(v) := Fa(v) + Fr(v)
attractive force (between adjacent vertices) P Fa(v) = c1 · (u,v)∈E log(duv)/c2)
repulsive force (between non adjacent vertices) P 1 Fr(v) = c3 · 2 u∈V duv
c1 = 2 c2 = 1 c3 = 1 c4 = 0.1
40 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
41 (Some notions of) Spectral graph theory
Laplacian matrix (simple graphs) G v1 v2 v4 deg(vi) if i = j QG[i, k] = v { −AG[i, j] otherwise v3 5
3 −1 −1 −1 0 −1 4 −1 −1 −1 QG = −1 −1 4 −1 −1 −1 −1 −1 4 −1 0 −1 −1 −1 3
42 (Some notions of) Spectral graph theory
Laplacian matrix (counting multiple edges) G v1 v2 v4 i = j deg(vi) if QG[i, k] = { −|edges| from vi to vj otherwise v3 v5
line i1, line i2, ···
QG[i1, i2, ...] = QG\ 0 { column i1, column i2, ··· G v0 1 v4 3 −1 −1 −1 0 −1 4 −1 −1 −1 5 −3 −2 v5 QG = −1 −1 4 −1 −1 Q 0 = G −3 4 −1 −1 −1 −1 4 −1 −2 −1 3 0 −1 −1 −1 3
43 (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 num- ber of spannig trees of G is: τ(G) = det(Q[i]) (i ≤ n)
G ··· v1 v 5 −3 −2 4 Q G −3 4 −1 4 −1 v5 −2 −1 3 Q [1] = G −1 3 = 11
44 First: existence and uniqueness of barycentric representations
45 First: existence and uniqueness of barycentric representations
Theorem Let G be a 3-connected planar graph with n vertices, and F a pe- ripheral 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 n x X ρ(v ) = w ρ(v ) i = 1,..., (n − k) { (I − W ) · y = by i ij j 1 (one equation for x = [x1, x2, . . . , xn−k] each inner vertex) X y = [y1, y2, . . . , yn−k] ρ(vi) = wijρ(vj) (coordinates of inner vertices) j∈N(i)
45 First: existence and uniqueness of barycentric representations Proof X ρ(vi) − wijρ(vj) = 0 j∈N(i)
(I − W ) · x = bx { (I − W ) · y = by
45 First: existence and uniqueness of barycentric representations Proof X ρ(vi) − wijρ(vj) = 0 j∈N(i)
(I − W ) · x = bx { (I − W ) · y = by
N(v ) = {v , v , v , v } 1 4 1 2 3 5 1 − x4 b4x v1 4 = 1 x b v1 − 5 5x v2 ρ(vi) := (xi, yi) 3 1 v4 1 1 − y4 b4y v3 v5 N(v5) = {v2, v3, v4} 4 = v2 −1 y5 b5y 3 1 v3
45 First: existence and uniqueness of barycentric representations Proof X X ρ(vi) − wijρ(vj) = 0 deg(vi)ρ(vi) − ρ(vj) = 0 j∈N(i) j∈N(i)
M · x = a (I − W ) · x = bx x M · y = a { (I − W ) · y = by { y
1 4 −1 1 − x4 b4x 4 = M = −1 3 1 x5 b5x −3 1 1 1 1 1 1 ρ(v4) − ρ(v5) = ρ(v1) + ρ(v2) + ρ(v3) 1 − y4 b4y 4 4 4 4 4 = 1 y b 1 1 1 − 5 5y { −3ρ(v4) + ρ(v5) = 3ρ(v2) + 3ρ(v3) 3 1
45 First: existence and uniqueness of barycentric representations
3 −1 −1 −1 0 4 −1 −1 4 −1 −1 −1 QG[1, 2, 3] := M = −1 3 QG = −1 −1 4 −1 −1
−1 −1 −1 4 −1 1 1 1 1 ρ(v4) − 4ρ(v5) = 4ρ(v1) + 4ρ(v2) + 4ρ(v3) 0 −1 −1 −1 3 1 1 1 { −3ρ(v4) + ρ(v5) = 3ρ(v2) + 3ρ(v3)
45 First: existence and uniqueness of barycentric representations
X M · x = ax deg(vi)ρ(vi) − ρ(vj) = 0 { M · y = a j∈N(i) y
G G/F det(M) = τ(QG/F ) > 0 G/F is connected
v1 G/F = G/{v1, v2, v3} v123 v v2 4 v4
v3 v5 v5 3 −1 −1 −1 0 4 −1 5 −3 −2 −1 4 −1 −1 −1 QG[1, 2, 3] := M = −1 3 QG/F −3 4 −1 Q = G −1 −1 4 −1 −1 −2−1 3 −1 −1 −1 4 −1 1 1 1 1 ρ(v4) − 4ρ(v5) = 4ρ(v1) + 4ρ(v2) + 4ρ(v3) 0 −1 −1 −1 3 1 1 1 { −3ρ(v4) + ρ(v5) = 3ρ(v2) + 3ρ(v3)
45 First: existence and uniqueness of barycentric representations
X M · x = ax deg(vi)ρ(vi) − ρ(vj) = 0 { M · y = a j∈N(i) y
G G/F det(M) = τ(QG/F ) > 0 G/F is connected M admits inverse
v1 G/F = G/{v1, v2, v3} v123 v v2 4 v4
v3 v5 v5 3 −1 −1 −1 0 4 −1 5 −3 −2 −1 4 −1 −1 −1 QG[1, 2, 3] := M = −1 3 QG/F −3 4 −1 Q = G −1 −1 4 −1 −1 −2−1 3 −1 −1 −1 4 −1 1 1 1 1 ρ(v4) − 4ρ(v5) = 4ρ(v1) + 4ρ(v2) + 4ρ(v3) 0 −1 −1 −1 3 1 1 1 { −3ρ(v4) + ρ(v5) = 3ρ(v2) + 3ρ(v3)
45 Second: the barycentric representation defines a planar drawing Theorem Let G be a 3-connected planar graph with n vertices, and F a pe- ripheral 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)
46 Second: the barycentric representation defines a planar drawing Theorem Let G be a 3-connected planar graph with n vertices, and F a pe- ripheral 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(v2) p2 ρ(v ) p 3 ρ(v ) 3 p ρ(v ) 1 1 4 ρ(v8) p4 p8 ρ(v) p ρ(v5) ρ(v7) p5 p7
ρ(v6) p6
barycentric representation of G planar drawing G
46 Second: the barycentric representation defines a planar drawing Theorem Let G be a 3-connected planar graph with n vertices, and F a pe- ripheral 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(v2) p2 ρ(v3) p3 ρ(v1) p1 ρ(v4) ρ(v8) p4 p8 ρ(v) p ρ(v7) p5 p7 ρ(v5)
ρ(v6) p6 σ(v) = α(v) σ(v) ≥ α(v) planar drawing of G dessin non planaire de G α(v) = 2π
46 Second: the barycentric representation defines a planar drawing Claim 1 σ(v) ≥ 2π = α(v)
ρ(v ) k 2 k ρ(v3) ρ(v1) ρ(v4) ρ(v8) γ i−1 γj β γ ρ(v) γi γi+1 ρ(v7) j δ ρ(v5) i j i ρ(v6) P σ(v) := k γk ≥ β + γ + δ = 2π
47 Second: the barycentric representation defines a planar drawing Claim 2 P P v α(v) ≤ v σ(v) = πf k
p2 γ p3 β + γ + δ = π p α 1 β δ p4 β p8 j γ η i p5 δ p7
p6 α + β + γ + δ + ... = (|F | − 2)π
inner vertices outer vertices P P P P v α(v) = α(v) + 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)
48 Second: the barycentric representation defines a planar drawing Conclusion α(v) = σ(v) Euler formula n − (e + |F |) + f = (|V \ F | + |F |) − (e + |F |) + (t + 1) 2π|V \ F | + (|F | − 2)π = πf inner edges inner triangles } Counts the number of edges 3t = 2e + |F |
P v α(v) := 2π|V \ F | + (|F | − 2)π = πf P P P P v α(v) = v σ(v) Claim 2 v α(v) ≤ v σ(v) = πf }
P P v α(v) = v σ(v) α(v) = σ(v) = 2π Claim 1 2π = α(v) ≤ σ(v) }
49 Third: peripheral cycles (non separating cycles)
Definition peripheral cycle v1 v2 A non-separating cycle C of a graph G is an in- v4 duced cycle in G such that G \ C is connected (after the deletion of edges and vertices of C) v3 v5
separating cycle v1 v2 A peripheral cycle C is a simple cycle of a con- v4 nected graph G such that for any two edges e1 v v5 and e2 in G \ C there exists a path (starting at 3 e1 and ending at e2) with no interior vertices on C no peripheral cycles
K2,3
50 Third: existence of peripheral cycles Lemma In a 3-connected planar graph peripheral cycles are exactly the faces (of the embedding) proof (exercise)
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