INF562, Lecture 3: Geometric and combinatorial properties of planar graphs
mardi 22 janvier 2013
Luca Castelli Aleardi
1-1 Intro Graph drawing: motivations and applications
2-1 Graph drawing and data visualization Global transportation system
3-1 Graph drawing and data visualization Roads, railways, ...
4-1 Graph drawing and data visualization Social network graph
5-1 Planar graphs french roads network Design of integrated circuits (VLSI)
6-1 Planar graphs french roads network Design of integrated circuits (VLSI)
www.2m40.com
9 accidents en 2012 (last one, on 28th sptember)
6-2 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
7-1 Mesh parameterization in geometry processing
8-1 Mesh parameterization in geometry processing
9-1 Mesh parameterization in geometry processing
10-1 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
11-1 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
11-2 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
11-3 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
11-4 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
11-5 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
11-6 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
11-7 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
11-8 Part I Major results in graph theory
12-1 Major results (on planar graphs) in graph theory
13-1 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
13-2 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
13-3 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
13-4 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).
13-5 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
13-6 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.
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) • µ(G) ≤ 3 λ1 = −4, λ2 = λ3 = λ4 = 0 v2 M ξx ξy ξz −1 −1 −1 −1 −1 −1 −1 −1 v0 −1 −1 −1 v3 −1 −1 −1 −1 v1 0 1 0 −1 −1 −1 −1 v2 0 0 1 v3 1 0 0 v0 v1
13-7 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
13-8 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)
13-9 Part II What is a surface mesh?
(a short digression on embedded graphs, simplicial complexes and topological and combinatorial maps)
14-1 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
15-1 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)`
16-1 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
17-1 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 un dessin planaire est un plongement cellulaire de G dans R2, qui satisfait les conditions suivantes: (i) les sommets du graphe sont reprsent´ es´ par des points ; (ii) les aretes sont reprsent´ ees´ par des arcs de courbes ne se coupant qu’aux sommets ;
(iii) les faces sont simplement connexes.
(topologica) map: cellular embedding up to homeomorphism (equivalence class)
two cellular embeddings defining the same planar map
18-1 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 goup 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) ...
19-1 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 goup 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) ...
19-2 Mesh representations: classification Manifold meshes non manifold or non orientable meshes
triangle meshes no boundary with boundaries genus 1 mesh Manifold mesh: definition Every edge is shared by at most 2 faces For every vertex v, the incident faces form an open or closed fan
v quad meshes polygonal meshes
20-1 Part III Euler formula and its consequences
21-1 Euler-Poincare´ characteristic: topological invariant χ := n − e + f
n − e + f = 2 Euler’s relation planar map
22-1 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
23-1 t Euler’s relation for polyhedral surfaces χ := n − e + f χ(t) = 3 − 3 + 2 = 2 First proof: by induction M P n − e + f = 2 Euler’s relation
χ(M) = χ(P ) − 1 (count exterior face)
M t
χ(M) = χ(M t) invariant: the boundary (exterior) is a simple cycle perform the removal according to a shelling order e000 = e00 − 2 f 000 = f 00 − 1 e0 = e − 1 f 0 = f − 1 e00 = e0 − 1 f 00 = f 0 − 1 n000 = n00 − 1
M t
remove a boundary edge remove a boundary edge remove a triangle
24-1 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
25-1 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
26-1 Euler’s relation for polyhedral surfaces Corollary: linear dependence between edges, vertices and faces f ≤ 2n − 4 e ≤ 3n − 6 preuve (par double comptage)
f = f1 + f2 + f3 + ... n = n1 + n2 + n3 + ... toutes les faces ont degre´ au moins 3 (G est simple), on a f = f3 + f4 + ... chaque arete apparait deux fois 2e = 3 · f3 + 4 · f4 + ... d’ou la relation 2e − 3f ≥ 0
27-1 Euler’s relation for polyhedral surfaces Corollary: linear dependence between edges, vertices and faces f ≤ 2n − 4 e ≤ 3n − 6 ayant prouve´ 2e − 3f ≥ 0
en appliquant Euler on trouve 3n − 6 = 3(e − f + 2) = 3e − 3f ≥ 0
28-1 Euler’s relation for polyhedral surfaces can we construct a regular mesh, where every vertex has degree 6?
29-1 Euler’s relation for polyhedral surfaces we just showed 2e − 3f ≥ 0
30-1 Major results (on planar graphs) in graph theory Kuratowski theorem (1930) (cfr Wagner’s theorem, 1937)
• G is planar iff it does not contain K5 nor K3,3 as minors
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
31-1 Part IV (Some notions of) Spectral graph theory
32-1 (Some notions of) Spectral graph theory
G v1 v2 v4
v3 v5
33-1 (Some notions of) Spectral graph theory
v adjacency matrix G 1 v2 v4 1 if vi is adjacent to vj AG[i, j] = v { 0 otherwise v3 5
0 1 1 1 0 1 0 1 1 1 AG = 1 1 0 1 1 1 1 1 0 1 0 1 1 1 0
33-2 (Some notions of) Spectral graph theory
v incidence matrix G 1 v2 v4 1 vi is incident to edge ek DG[i, k] = v { 0 otherwise v3 5
e1 e1 e2 e3 e4 e5 e6 e7 e8 e9
0 0 1 1 1 0 0 0 0 v1 e3 e4 1 1 1 0 0 1 0 0 0 v2 e5 D = ...... G e6 ...
e2
33-3 (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
33-4 (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
33-5 (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
34-1 Part V Tutte’s planar embedding
35-1 Preliminaries: barycentric coordinates Pn P q = i αipi (avec i αi = 1)
coefficients (α1, . . . , αn) are called barycentric coordinates of q (relative to p1, . . . , pn) w = 0
p2 (u, v, w) = (0, 1, 0) q u < 0, v > 0, w > 0
p3 p1 (0, 0, 1) (1, 0, 0) v = 0
u > 0, v < 0, w > 0 u = 0
36-1 Tutte’s theorem Thm (Tutte barycentric method, 1963) Every 3-connected planar graph G admits a convex representation ρ in R2.
planar drawing of G two straight-line planar drawings of G v1 v v2 v1 1 v4 v1 v3 v2 v4 v2 v3 v5 v2 v3 v5 v3
37-1 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 est convexe the images of the faces of G are convex polygons
planar drawing of G two straight-line planar drawings of G v1 v v2 v1 1 v4 v1 v3 v2 v4 v2 v3 v5 v2 v3 v5 v3
37-2 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} planar drawing of G two straight-line planar drawings of G v1 v v2 v1 1 v4 v1 v3 v2 v4 v2 v3 v5 v2 v3 v5 v3
37-3 Tutte’s theorem: main steps v1 find a peripheral cycle F (the outer face of G) v2 v4 a cycle such that G \ F is connected v3 v5 (deletion of vertices and edges)
38-1 Tutte’s theorem: main steps v1 find a peripheral cycle F (the outer face of G) v2 v4 a cycle such that G \ F is connected v3 v5 (deletion of vertices and edges)
v1 choose a convex polygon P of size k = |F | v such that ρ(F ) = P 2
v3
38-2 Tutte’s theorem: main steps v1 find a peripheral cycle F (the outer face of G) v2 v4 a cycle such that G \ F is connected v3 v5 (deletion of vertices and edges)
v1 choose a convex polygon P of size k = |F | v such that ρ(F ) = P 2
v3
solve equations for images of inner vertices ρ(vi):
X X ρ(vi) = wijρ(vj) ρ(vi) − wijρ(vj) = 0 j∈N(i) j∈N(i) 1 according to Tutte: wij = deg(vi)
38-3 Tutte’s theorem: main steps v1 find a peripheral cycle F (the outer face of G) v2 v4 a cycle such that G \ F is connected v3 v5 (deletion of vertices and edges)
v1 choose a convex polygon P of size k = |F | v such that ρ(F ) = P 2
v3
solve a linear system: 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)
38-4 Tutte’s theorem: main steps v1 find a peripheral cycle F (the outer face of G) v2 v4 a cycle such that G \ F is connected v3 v5 (deletion of vertices and edges)
v1 choose a convex polygon P of size k = |F | v such that ρ(F ) = P 2
v3
solve a linear system: N(v4) = {v1, v2, v3, v5} N(v5) = {v2, v3, v4} 1 1 − x4 b4x ρ(vi) := (xi, yi) 4 = 1 x5 b5x −3 1 ρ(v ) − 1ρ(v ) = 1ρ(v ) + 1ρ(v ) + 1ρ(v ) 1 1 y b 4 4 5 4 1 4 2 4 3 −4 4 4y 1 = 1 1 1 y5 b5y { −3ρ(v4) + ρ(v5) = 3ρ(v2) + 3ρ(v3) −3 1
38-5 Validity of Tutte’s theorem: main results show that the linear system admit a (unique) solution: (I − W ) · x = bx 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 Exercice Claim (existence of peripheral cycles) v2 v4 In a 3-connected planar graph peripheral cycles are exactly the faces (of the embedding) v3 v5
39-1 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
(I − W ) · x = bx linear systems to solves { (I − W ) · y = by
nice drawings (detection of symmetries)
40-1 Drawbacks of Tutte’s drawing
requires to solve linear systems of equations (of size n) (I − W ) · x = bx complexity O(n3) or O(n3/2) with methods more involved { (I − W ) · y = by
exponential size of the resulting vertex coordinates (with respect to n)
drawings are not always ”nice”
41-1 Tutte’s spring embedder: iterative version F P choose an outer face , and a convex polygon P put exterior vertices v ∈ F on the polygon F(v) = (u,v)∈E(pu − pv) 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
42-1 Spring drawing
placer tous les points aleatoirement´ dans le plan repeter (jusqu’a` convergence) pour tout sommet v ∈ V calculer
v = v + c4 · F(v)
ou` F(v) := Fa(v) + Fr(v)
force attractive (entre sommets voisins) P Fa(v) = c1 · (u,v)∈E log(dist(u, v)/c2) (u, v) arete reliant le sommet v a` u
P 1 Fr(v) = c3 · √ u∈V dist(u,v) force repulsive (entre tous les sommets)
c1 = 2 c2 = 1 c3 = 1 c4 = 0.01
43-1 Part VI Tutte’s theorem: the proof
44-1 First: existence and uniqueness of barycentric representations
45-1 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 the the systems above admit a solution (unique)
(I − W ) · x = bx X ρ(vi) − wijρ(vj) = 0 { (I − W ) · y = by j∈N(i)
45-2 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) the linear systems above are equivalent (I − W ) · x = bx M · x = ax { (I − W ) · y = by { M · y = ay
45-3 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) the linear systems above are equivalent (I − W ) · x = bx M · x = ax { (I − W ) · y = by { M · y = ay
1 1 − x4 b4x 4 = −1 x5 b5x 3 1 4 −1 1 1 − y4 b4y M = −1 3 4 = 1 y5 b5y −3 1
45-4 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) the linear systems above are equivalent (I − W ) · x = bx M · x = ax { (I − W ) · y = by { M · y = ay
4 −1 1 3 −1 −1 −1 0 1 − x4 b4x 4 = −1 3 1 x b −1 4 −1 −1 −1 − 5 5x QG[1, 2, 3] = M 3 1 QG = −1 −1 4 −1 −1 1 1 − y4 b4y 4 = −1 −1 −1 4 −1 laplacian matrix 1 y5 b5y −3 1 0 −1 −1 −1 3
45-5 First: existence and uniqueness of barycentric representations Proof X M · x = ax deg(vi)ρ(vi) − ρ(vj) = 0 { M · y = a j∈N(i) y
3 −1 −1 −1 0 −1 4 −1 −1 −1 QG = −1 −1 4 −1 −1 4 −1 −1 −1 −1 −1 4 −1 3
0 −1 −1 −1 3 QG[1, 2, 3] = M
45-6 First: existence and uniqueness of barycentric representations Proof 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 v2 v123 v v4 G/F = G/{v1, v2, v3} 4
v3 v5 v5 3 −1 −1 −1 0 QG/F −1 4 −1 −1 −1 5 −3 −2 QG = −1 −1 4 −1 −1 4 −1 edge contraction −3 4 −1 −1 −1 −1 −1 4 −1 3 −2 −1 3 0 −1 −1 −1 3 QG[1, 2, 3] = M
45-7 First: existence and uniqueness of barycentric representations Proof conclusion 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 v2 v123 v v4 G/F = G/{v1, v2, v3} 4
v3 v5 v5 3 −1 −1 −1 0 QG/F −1 4 −1 −1 −1 5 −3 −2 QG = −1 −1 4 −1 −1 4 −1 edge contraction −3 4 −1 −1 −1 −1 −1 4 −1 3 −2 −1 3 0 −1 −1 −1 3 QG[1, 2, 3] = M
45-8 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)
46-1 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
46-2 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
46-3 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
47-1 Second: the barycentric representation defines a planar drawing Claim 2 P P v α(v) ≤ v σ(v) = πf k
p2 γ p 3 p α 1 β + γ + δ = π α + β + γ + δ + ... = (|F | − 2)π β δ p4 β p8 j γ η i p5 δ p7
p6
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)
48-1 Second: the barycentric representation defines a planar drawing Conclusion α(v) = σ(v) P P Claim 2 v α(v) ≤ v σ(v) = 2π
n − (e + |F |) + f = (|V \ F | + |F |) − (e + |F |) + (t + 1) Euler formula 3t = 2e + |F | } Counts the number of edges
P P v α(v) := 2π|V \ F | + (|F | − 2)π = πf = v σ(v)
P P v α(v) = v σ(v) α(v) = σ(v) = 2π Claim 1 2π = α(v) ≤ σ(v) }
49-1 The missing proof 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
50-1 The missing proof Lemma (Laplacian and the number of spanning trees)
51-1 The missing proof Lemma (Laplacian and the number of spanning trees)
Claim 1 Pour toute arete e de G on a:
τ(G) = τ(G/e) + τ(G \ e) id´eede la preuve
tout arbre couvrant de G ne contenant pas e est aussi un arbre cou- vrant de G \ e: il y en a τ(G \ e) il y a une correspondance bijective entre les arbres couvrants de G est les arbres couvrants de G/e
51-2 The missing proof Lemma (Laplacian and the number of spanning trees) Claim 2 Consid´erons une arete e = (u, v) et la matrice E ci-dessous: v v4 v3 QG 0 E 0 3 −1 −1 −1 1 v G −1 2 −1 0 −1 −1 3 −1 0 0 −1 0 −1 2 v1 v2 on a: QG[u] = QG\e + E e = (v3, v4)
QG\e[v3] E 3 −1 −1 −1 3 −1 −1 −1 0 QG[v3] −1 2 −1 0 −1 2 −1 0 0 −1 −1 3 −1 −1 −1 2 0 0 −1 0 −1 2 −1 0 0 1 1
51-3 The missing proof Lemma (Laplacian and the number of spanning trees) Claim 2
observons que: v4 v3 QG QG\e[u, v] = Q[u, v] 3 −1 −1 −1 G −1 2 −1 0 −1 −1 3 −1 −1 0 −1 2 v1 v2 detQ[u] = detQG\e[u] + detQG\e[u, v]
e = (v3, v4)
QG\e[v3] QG\e[v3, v4] 3 −1 −1 −1 3 −1 −1 −1 3 −1 −1 −1 QG[v3] −1 2 −1 0 −1 2 −1 0 −1 2 −1 0 −1 −1 3 −1 −1 −1 2 0 −1 −1 2 0 −1 0 −1 2 −1 0 0 1 −1 0 0 1
51-4 The missing proof Lemma (Laplacian and the number of spanning trees)
Claim 4 3 −1 −1 −1 −1 2 −1 0 v4 v3 QG v4 −1 −1 3 −1 G/e −1 0 −1 2 G
v1 v2 v1 v2
QG/e[v] = Q[u, v]
3 −1 −1 −1 3 −1 −2 QG[v3, v4] −1 2 −1 0 −1 2 −1 QG/e[v4] −1 −1 3 −1 −2 −1 3 −1 0 −1 2
51-5 The missing proof Lemma (Laplacian and the number of spanning trees)
Fin: on utilise l’induction 3 −1 −1 −1 v v 4 3 v4 −1 2 −1 0 QG G/e −1 −1 3 −1 G −1 0 −1 2
v1 v2 v1 v2
detQ[u] = detQG\e[u] + detQG\e[u, v]
detQ[u] = detQG\e[u] + detQG/e[v] τ(G) = detQ[u]
τ(G \ e) τ(G/e)
51-6