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Planar Orientations

Chapter 4 (4.1-4.6) in the bookWritten By: Tomer Heber

Outline & Goal• In this lesson we will deal with planar st-graphs (which will be

explained later on).• We will review several algorithms for representing a st-planar

graph.• We will also learn that by using those representations we can

turn a st-planar graph, to a more specific shape (such as a polyline).

The goal of this lesson is simple: Which is to learn of ways to represent a st-graph in a more “useful” and “aesthetic” representation.

Numbering of Digraphs

• A topological numbering of G is an assignment of numbers to the vertices of G, such that, for every edge (u,v) of G, the number assigned to v is greater then the one assigned to u.

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Numbering of Digraphs (2)

• A topological sorting is a topological numbering of G, such that every vertex is assigned a distinct integer between 1 and n.

• A topological sorting is unique if G has a directed path that visits every vertex.

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Numbering of Digraphs (3)

• The following statements are equivalent:

• G is acyclic.• G admits a topological numbering.• G admits a topological sorting.

In other words a topological numbering (sorting) can be done only on an acyclic graph.

Planar Graph

• A planar graph is a graph which can be embedded in the plane, i.e., it can be drawn on the plane in such a way that its edges intersect only at their endpoints.

Faces

ST-Graph

• An acyclic digraph with a single source s and a single sink t is called an st-graph.

t

s

Sink “t” has no outgoing edges!

Source “s” has no incoming edges!

ST-Graph (2)

• Let G be an st-graph. The following simple properties hold:

• Given a topological numbering of G. every directed path of G visits with increasing numbers.

• For every vertex v of G, there exists a simple directed path from s (source) to t (sink) that contains v.

Planar ST-Graph

• A planar st-graph is an st-graph that is planar and embedded with vertices s and t on the boundary of the external face.

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Planar ST-Graph (2)

• Let G be a planer st-graph and F be its set of faces. We conventionally assume that F contains two representatives for the external face:

• The “left external face” s*• The “right external face” t*

s* t*

Planar ST-Graph (3)

• For each edge e = (u,v), we define orig(e) = u and dest(e) = v.

• We define left(e) (resp. right(e)) to be the face to the left (resp. right) of e.

eu

vf1

f2

orig(e) = udest(e) = v

left(e) = f1left(e) = f2

Planar ST-Graph Properties

• Lemma 1: Each face f of G consists of two directed paths with common origin called orig(f), and common destination called dest(f).

• Proof: Let f be a face of G for which the lemma is not true.

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There must be a path from vertex u to the “sink” t

There must be a path from the “source” s to vertex w

Since the graph is planar there must be a node where the passes intersectWe receive a cycle!

Planar ST-Graph Properties (2)

• Lemma 2: The incoming edges for each vertex of G appear consecutively around v, and so do the outgoing edges.

• Proof: The lemma holds trivially for the vertices s and t. Let v be any other vertex, and suppose for a contradiction, that there are edges (v,w0), (w1,v), (v,w2), (w3,v).

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w3w2

w1w0

Because every vertex in a st-graph has a simple path from “s” to “t” we add the following paths…Because the graph is planar we add the vertex “x” to the graph.

xWe now have a cycle!

Planar ST-Graph Properties (3)

• Since the incoming (outgoing) edges of each vertex of G appear consecutively we define the face separating the incoming edges from the outgoing edges in clockwise order, left(v), and the other separating face is called right(v).

Planar ST-Graph G*

• We define a digraph G* associated with planar st-graph G, as follows:

• The vertex set of G* is the set F of faces (recall that F has two representatives, s* and t*, of the external face.• For every edge e != (s,t) of G, G* has an e* = (f,g) where

f = left(e) and g = right(e).

eleft(e) right(e)

Notice that G* is a planar st-graph as well

Lemma 3

• An element of is called an object of planar st-graph G.

• For vertex v, we define orig(v) = dest(v) = v.

• For a face f, we define left(f) = right(f) = f.

* Reminder we have already defined previously: left(v), right(v), left(e), right(e), orig(e), dest(e), orig(f) and dest(f).

Lemma 3 (2)

• Lemma 3: For any two objects o1 and o2 of a planar st-graph G, exactly one of the following holds:

• G has a directed path from dest(o1) to orig(o2).

• G has a directed path from dest(o2) to orig(o1).

• G* has a directed path from right(o1) to left(o2).

• G* has a directed path from right(o2) to left(o2).

Tile

• A tile is a rectangle with sides parallel to the coordinate axes.

• A tile can be unbounded or can degenerate to a segment or a point.

• Two tiles are horizontally (vertically) adjacent if they share a portion of a vertical (horizontal) side.

Tiles: Vertically adjacent:

Tessellation Representation

• Let G be a planar st-graph. A tessellation representation Θ for G maps each object o of G into a tile Θ(o) such that:

• The interiors of tiles Θ(o1) and Θ(o2) are disjoint whenever o1 != o2.

• The union of all tiles Θ(o), is a rectangle.• Tiles Θ(o1) and Θ(o2) are horizontally adjacent if and only if o1 =

left(o2) or o1 = right(o2) or o2 = left(o1) or o2 = right(o1).• Tiles Θ(o1) and Θ(o2) are vertically adjacent if and only if o1 =

orig(o2) or o1 = dest(o2) or o2 = orig(o1) or o2 = dest(o1).

FEVo

Tessellation Representation (2)

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X(left(e)) = 0X(right(e)) = 1Y(orig(e)) = 0Y(dest(e)) = 2

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X(left(f)) = 3X(right(f)) = 3Y(orig(f)) = 2Y(dest(f)) = 4

Tessellation Representation (3)

• The correctness of the algorithm is based on Lemma 3:• Let there be tile t1 and tile t2, from Lemma 3 t1 is either: above t2,

below t2, left of t2 or right of t2. And only one of this directions is true.

• Since each line of the algorithm is O(n), the total runtime of the algorithm is O(n).

• The size of the Tessellation Representation can be modified by modifying the topological numbering (e.g. increasing the numbering to be 0..2..4 instead of 0..1..2 will make a Tessellation Representation twice bigger).

Visibility Representation

• Let G be a planar st-graph. A visibility representation of G draws each vertex v as a horizontal segment, called vertex segment , and each edge (u,v) as vertical segment, called edge segment such that:

• The vertex segments do not overlap.• The edge segments do not overlap.• Edge-segment has its bottom end point

on , its top end-point on , and does not intersect any other vertex segment.

)(v

),( vu

),( vu )(u)(v

Visibility Representation (2)


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• The correctness of the algorithm By lemma 3 and the construction of the algorithm:

• Any two vertex segments are separated by a horizontal or vertical strip of at least unit width (The vertex segments do not overlap).

• Any two edge segments on opposite sides of a face are separated by a vertical strip of at least a unit width (The edge segments do not overlap).

• Each edge segments (u,v) has its bottom point intersecting with u vertex segment, and his upper point intersecting with v vertex segment (sufficing the 3rd condition).

• The runtime of the algorithm is O(n) since each step is O(n).

Constrained Visibility Representation

• Let G be a planar st-graph with n vertices. Two paths of π1 and π2 of G are said to be nonintersecting if:

• They are edge disjoint.• And do not cross at common vertices.

π1

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Constrained Visibility Representation (2)

• Given a collection ∏ of nonintersecting paths of G, we consider the problem of constructing a visibility representation Γ of G such that for every path π in ∏ we have the following constraint:

• Any two edges e’ and e’’ of π, the edge segments Γ(e’) and Γ(e’’) have the same x-coordinate.

Note: We assume that ∏ covers all the edges in graph G (Such that any edge who is not in any of the nonintersecting paths in the collection ∏, is a nonintersecting path himself in ∏).


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∏={π1, π2,π3,π4,π5, π6, π7, π8}

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Polyline Drawing

• We can construct a planar upward polyline drawing of a planar st-graph G using its visibility representation.

• We draw each vertex in an arbitrary point inside its vertex segment.

• We draw each edge (u,v) of G as a three segment polygonal chain.

Polyline Drawing (2)

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Polyline Drawing (3)

• Since every step in the algorithm is O(n), the total complexity runtime is O(n).

• We can reduce the number of “bends” if we put each place vertex on intersections between an edge segment and vertex segment.

Polyline Drawing (4)• The technique can be extended for constrained visibility representation of

a planar st-graph which is called constrained-polyline.• In a constrained-polyline all the internal vertices in a path π in ∏ are

vertically aligned.

The algorithm forConstrained polyline

QUESTIONS???