Computational Geometry Seminar Lecture 1 Drawing Planar Graphs

Computational Geometry Seminar Lecture 1

Drawing Planar Graphs

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Graph

• Definition: A Simple Undirected Graph G consists of:

A finite set of vertices V(G).

A set E(G) of sets of 2 vertices, called edges.

• Note: From now on we will call this type of graphs by the general name Graph.

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Degree

• Definition: In a graph G, the vertices u, v are adjacent iff the edge uv belongs to E(G).

• The number of adjacent vertices to a vertex u is called the degree of u, denoted d(u).

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Subgraph

• Definition: A graph H is a subgraph G (written H G) iff V(H) V(G) and E(H) E(G).

• We say that a graph H is the subgraph of G induced by a set of vertices U V(G) if V(H) = U and E(H) is the set of all the edges of G connecting vertices of U.

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Paths and Cycles

• Definition: A sequence of k distinct vertices, in which every consecutive vertices are adjacent, is called a path of length k-1.

• Definition: A path of length k-1 with the addition of an edge from the last vertex to the first vertex of the path is called a cycle of length k.

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Connectivity

• Definition: A maximal set of vertices, for which there exists a path from every vertex in the set to another, is called a connected component.

• A graph composed of only one connected component is called connected.

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Graphs in the plane

• Can represent graphs in the plane by assigning distinct points to vertices and drawing continuous non-self-intersecting curves (Jordan arcs) between adjacent vertices.

K4

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Graphs in the plane

• But sometimes we want the drawing to be simple or satisfy other requirements, such as: straight line segments as arcs or avoiding crossing arcs.

K4

Can we always achieve that?

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Planar Graphs

• Definition: A graph that can be represented in the plane so that no two arcs meet at a point other than their endpoints is called a Planar graph.

• Such representation of a planar graph is called a Plane graph or Planar embedding of the graph.

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Nonplanar Graphs

K5 K3,3

K5 and K3,3 are not planar

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K3,3 is not planar

v3

u1

v2

u2

v1

u3

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Subdivision

• A subdivision of a graph is obtained by repeating the operation of removing an edge and introducing a new vertex connected to the endpoints of the edge removed.

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Kuratowski’s Theorem

• The theorem states that a graph is not planar iff it has a subgraph which is a subdivision of K5 or K3,3.

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Straight Line Embedding

• Deleting any edge from K5 will result in a planar graph. Moreover this graph can be embedded in the plane by using straight line segments.

• Does every planar graph have a straight line embedding???

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Faces

• Definition: A plane graph divides (with its arcs) the plane into connected regions called faces.

• Exactly one of these faces is unbounded and is called the exterior face.

• We denote the number of faces of a plane graph G by f(G).

Exterior Face

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Dual Graph

• Definition: For a plane graph G we construct G*, the dual of G as follows. A vertex is placed in each face of G. These are the vertices of G*. For each edge e of G we draw an edge e*, called the dual edge of e, which crosses e (and no other edge of G) and joins the vertices corresponding to the faces, whose boundary consists of e.

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Euler’s Formula

• For a connected plane graph G:

v( ) - e( ) + f( ) = 2G G G

v = 5

e = 7

f = 4

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Euler’s Formula (Proof)

• Proof: By induction on f. If f(G) = 1 then G has no cycles (a tree), thus

e(G) = v(G) – 1.

Assume for f(G) ≥ 2 that the theorem is correct for connected plane graphs with fewer than f(G) faces.

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Euler’s Formula (Proof)

Delete an edge e that belongs to a cycle in G. For the resulting connected plane graph G-e we get f(G-e) = f(G) – 1. Therefore, using the induction hypothesis on G-e we obtain:

v(G) – (e(G) – 1) + (f(G) – 1) = 2. ☺

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Euler’s Formula

• Note: When dealing with disconnected graphs we can add edges between connected components without adding more faces, eventually creating a connected graph. Thus, the formula becomes: (C is the number of connected components)

v(G) - e(G) + f(G) = 1 C

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Bridges

• Definition: An edge which is a boundary to only one face is called a bridge.

• Same bridges as in graph theory (edges not contained in cycles). Bridges

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Sides

• Definition: For a face f of G, the number of sides of f is the number of edges belonging to the boundary of f, where bridges are counted twice. Denoted s(f).

5

7

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Sides

• Every non-bridge is in a boundary of exactly two faces. Therefore:

( )

( ) 2 ( )f F G

s f e G

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Euler’s Formula

• Note: According to Euler’s formula the number of faces is independent of the embedding we choose for the graph. However, the number of sides of the faces is not.

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Triangulation

• Definition: A face f for which s(f) = 3 is called a triangle. If all faces of G are triangles, G is called a Triangulation.

TriangleTriangulation

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Triangulation

• Every graph can be extended to a triangulation by the addition of new edges between existing vertices.

• A triangulation is maximal in the sense that no more edges can be added without violating its planarity.

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Euler’s Formula

• Corollary 1: For a plane graph G with at least 3 vertices:

e(G) ≤ 3v(G) – 6

f(G) ≤ 2v(G) – 4

• The equalities hold iff G is a triangulation.

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Corollary 1

• Proof: Sufficient to prove for connected plane graphs since otherwise number of edges and faces only decreases.

For every face f of G, s(f) ≥ 3.Thus:

By Euler’s formula we obtain

v(G) – e(G) + e(G) ≥ 2

v(G) – f(G) + f(G) ≥ 2 ☺

( )

3 ( ) ( ) 2 ( )f F G

f G s f e G

2

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2

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Chromatic Number

• Definition: The chromatic number χ(G) of a graph G is the minimum number of colors required to color the vertices of G so that no adjacent vertices are of the same color.

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The Four-Color Theorem

• According to the four-color-theorem of Appel and Haken the chromatic number of a planar graph is at most 4.

• This bound cannot be improved.

• Proof is quite complicated. We’ll prove a weaker statement deduced from Corollary 1.

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Corollary 2

• Corollary 2: If G is a planar graph, then χ(G) ≤ 5.• Proof: By induction on v(G). If v(G) ≤ 5, we can assign every vertex a different

color.Assume that for v(G) ≥ 6 we proved the statement for

graphs of size smaller than v(G).

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Corollary 2 (Proof)

From Corollary 1: G must have a vertex u with d(u) ≤ 5.

Otherwise for every vertex u, d(u) ≥ 6. Thus:

( )

( )6 ( )

( ) 3 ( )2 2

u V G

d uv G

e G v G

In contradiction to corollary 1.

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Corollary 2 (Proof)

If d(u) ≤ 4 then color the rest of the graph G-u with 5 colors using the induction, and then color u with a color different from its neighbors.

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Corollary 2 (Proof)

If u has 5 neighbors wi (1≤ i ≤ 5):Since G is planar it does not contain K5 as a subgraph. Thus, assume WLOG that w1 and w2 are not adjacent.

Let G’ be the graph obtained from G-u by merging w1 and w2 to w’, which is adjacent to neighbors of w1 or w2.

Merge

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Corollary 2 (Proof)

G’ is a planar graph, hence we apply the induction to obtain a 5-coloring of G’. If we use the same coloring on G-u, where w1 and w2 are assigned the color of w’, the neighbors of u are colored in at most 4 colors. Therefore, we can assign a color for u different from its neighbors. ☺

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Straight Line Drawing

• We will now prove that every planar graph has and embedding with straight line segments, called the straight-line embedding.

K4

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u1


• Let us start off with the following lemma:• Lemma 1: Let G be a plane graph whose exterior

face is bounded by a cycle u1, … , uk. Then exists up (p ≠ 1, k) not adjacent to any uj other than up-1 and

up+1.

u2

u3

u4 u5

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u1

Lemma 1 (Proof)

• Proof: If no two non-consecutive vertices in the exterior boundary are adjacent, then the lemma is trivial. Otherwise:

Pick two non-consecutive adjacent vertices ui, uj (j > i+1) for which j-i is minimal.

u2

u3

u4 u5

i = 1

j = 3

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u1

Lemma 1 (Proof)

ui+1 cannot be adjacent to u1, …, ui-1, …, uj+1 , …, uk by planarity.

u2

u3

u4 u5

crossing

ui+1 cannot be adjacent to any ui, …, uj other than ui and ui+2 because of minimality of j-i. ☺

i = 1

j = 3

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Canonical Construction of Triangulations

• Let G be a triangulation with an exterior face uvw and a labelling u1 = u, u2 = v, u3, …, un = w of the vertices of G. Denote by Gk the subgraph of G induced by u1, …, uk and by Ck the exterior boundary of Gk.

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Canonical Construction of Triangulations

• There exists a labelling such that for every 4 ≤ k ≤ n:

I. Gk-1 is internally triangulated.

II. The edge uv is in Ck-1.

III. uk is in the exterior face of Gk-1 the neighbors of uk in V(Gk-1) are consecutive on Ck-1.

• This kind of labelling is called a canonical labelling.

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Canonical Construction of Triangulations (Proof)

• Proof: We define un, un-1, …, u3 by reverse induction.For un = w:

u v

w

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Canonical Construction of Triangulations (Proof)

For 4 ≤ k ≤ n: Assume we defined un, …, uk correctly. Applying Lemma 1 to Gk-1, the subgraph induced by the remaining vertices, we know that there is a vertex on Ck-1, other than u and v, that is only adjacent to its preceding and subsequent vertices on Ck-1. Let uk-1 be that vertex. Proof of I.-III. is the same as with un.

☺

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Canonical Labelling (Example)

u v

w

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u= u1u2=v

w= u7

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u= u1u2=v

u6

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u= u1u2=v

u5

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u= u1u2=v

u4

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u= u1u2=v

u3

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• Corollary 3: Every planar graph has a straight line embedding in the plane.

• Proof: It is sufficient to show that the statement is true for any maximal planar graph, i.e. a graph that can be represented as a triangulation.

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Corollary 3 (Proof)

Let G be a triangulation with the canonical labelling u1 = u, u2 = v, u3, …, un = w as described earlier. We will determine the positions p(uk) = (x(uk), y(uk)) of the vertices by induction on k.

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Corollary 3 (Proof)

Set p(u1) = (0,0) , p(u2) = (2,0) , p(u3) = (1,1).

For k ≥ 4 assume that p(u1), …, p(uk-1) have already been defined s.t., by connecting the images of adjacent vertices with segments, we obtain a straight line embedding of Gk-1, whose exterior face is bounded by the segments corresponding to the edges of Ck-1.

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Corollary 3 (Proof)

Let v1 = u, v2, …, vm = v denote the vertices of Ck-1 listed in the order they appear on the cycle.Suppose further that:

(1) x(v1) < x(v2) < … < x(vm)y(vi) > 0 for 1 < i < m

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Corollary 3 (Proof)

By condition III. of the canonical labelling, uk is connected to a series of consecutive vertices of Ck-1, vq, …, vr (1 ≤ q < r ≤ m). Choose x(uk) s.t. x(vq) < x(uk) < x(vr). Now choose y(uk) > 0 large enough. The new position p(uk) meets all the requirements (including the assumption (1)). ☺

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Straight Line Drawing (Example)

The starting graph

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Making the triangulation

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u= u1

v= u2

w= u6

u5 u4

u3

Canonical labelling

58


u= u1

v= u2

w= u6

u5u4

u3

Left:

0 1 2

x

Setting p(u1) p(u2) p(u3)

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u= u1

v= u2

w= u6

u5u4

u3

Left:

0 1 2

xv1

v2

v3

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u= u1

v= u2

w= u6

u5u4

u3

Left:

0 1 2

xv1

v3

v4

v2

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u= u1

v= u2

w= u6

u5u4

u3

Left:

0 1 2

xv1

v3

v4

v2

v5

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Delete added edges

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• Note: By using this method we can also establish the existence of straight line embedding of other representations. For example, straight line embeddings of the following representation, found by Rosenstiehl and Tarjan.

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• Corollary 4: The vertices and edges of any planar graph can be represented by horizontal and vertical segments, respectively, s.t.:I. Segments don’t have interior points in common.II. Horizontal segments connected by a vertical segment represent adjacent vertices.

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Corollary 4 (Proof)

• Proof: As earlier it is sufficient to establish the statement for triangulations.

Let G be a triangulation with the canonical labelling u1 = u, u2 = v, u3, …, un = w. To every uk we will assign a segment s(uk) whose endpoints are (xk, k) and (x'k, k).

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Corollary 4 (Proof)

Set x1=0, x'1=2, x2=2, x'2=4, x3=1, x'3=3.

For k ≥ 4 assume that s(u1), …, s(uk-1) have already been defined s.t., the subgraph Gk-1 has a representation satisfying conditions I. and II.

0 2 4

x

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Corollary 4 (Proof)

We define the upper envelope of s(u1), …, s(uk-1) as the set of points for which no other point is directly above them. Denote this by UEk-1.

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Corollary 4 (Proof)

Let v1 = u, v2, …, vm = v denote the vertices of Ck-1 listed in the order they appear on the cycle.Suppose further that UEk-1 consists of portions of s(v1), …, s(vm), in this order (from left to right).

s(v1)

s(v3)s(v2)

s(v4)

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Corollary 4 (Proof)

By condition III. of the canonical labelling, uk is connected to a series of consecutive vertices of Ck-1, vq, …, vr (1 ≤ q < r ≤ m). Choose xk to be some point of s(vq) which belongs to UEk-1 and choose x'k to be some point which belongs to UEk-1. The resulting representation of Gk meets the requirements. ☺

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u= u1

v= u2

w= u6

u5 u4

u3

Canonical labelling

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u= u1

v= u2

w= u6

u5u4

u3

Left:

0 2 4

x

Setting s(u1) s(u2) s(u3)

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u= u1

v= u2

w= u6

u5u4

u3

Left:

0 2 4

xs(v1)

s(v3)

s(v2)

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u= u1

v= u2

w= u6

u5u4

u3

Left:

0 2 4

xs(v1)

s(v3)s(v2)

s(v4)

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u= u1

v= u2

w= u6

u5u4

u3

Left:

0 2 4

xs(v1)

s(v3)

s(v2)

s(v4)

s(v5)

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Delete added edges

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3D Implementations

• So far we’ve seen embeddings of graphs in 2D (on the plane).

• We will now see how to embed planar graphs in 3D.

• We will also get to know new shapes which allow use to draw graphs which are not planar.

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The Sphere

• Every planar graph can be embedded on the sphere.

• Likewise, every embedding on the sphere can be transformed to an embedding on the plane.

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The Sphere

• To transform an embedding on the sphere to an embedding on the plane:

• We choose a face and puncture a hole in it. Then stretch the face using the hole until we get a flat surface. That face will be the exterior face.

• This procedure is reversible.

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Handle

• Definition: A handle is a tube joining two holes cut in a surface.

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Genus

• Definition: The genus of a surface obtained by adding handles to a sphere is the number of spheres added.

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Genus

Genus 0: Sphere Genus 1: Torus

Genus 2: Double-Torus Genus 3: Pretzel

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Graphs

• The genus of a graph is the minimum γ such that the graph can be embedded on a surface of genus γ.

• As stated planar graphs can be embedded on the sphere. Therefore, they are of genus 0.

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Graphs

• K5 and K3,3 are of genus 1.

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Computational Geometry Seminar Lecture 1 Drawing Planar Graphs