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Orthogonal Drawing (continued)

Sections 8.3 – 8.5 from the book

Bert Spaan

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Orthogonal Drawing (continued)

1. Linear Algorithm Without Bad Cycles With Bad Cycles

2. Orthogonal Grid Drawing

3. Orthogonal Drawings without Bends

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Linear Algorithm for Bend-Optimal Drawing

Linear time algorithm for graphs with the following properties: 3-connected Δ ≤ 3 Four or more outer edges

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k-Connectivity

A graph G is connected if for any distinct vertices u and v there is a path between u and v in G.

A graph is k-connected if the minimum number of vertices whose removal results in a disconnected graph is k.

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Examples

1-connected:

2-connected:

3-connected:

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Bad Cycles A graph has no bad cycles and thus a

rectangular drawing if and only if Any 2-legged cycle contains two or more

corners Any 3-legged cycle contains one or more

corners A bad cycle is a 2-legged cycle which

contains at most one corner, or a 3-legged cycle which contains no corners.

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Maximal Bad Cycles A bad cycle C in G is defined to be maximal

if C is not contained in any other bad cycle in G

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Linear AlgorithmAlgorithm Orthogonal-Draw(G)begin Add four dummy vertices of degree two on four distinct outer edges as the corners; Let G’ be the resulting graph; Let C1, C2 , …, Cl, be the maximal bad cycles in G’;

for each i, 1 ≤ i ≤ l, construct a genealogical tree TCi and determine green paths

and red paths for every cycle in TCi;

for each i, 1 ≤ i ≤ l, find an orthogonal drawing D(G(Ci)) of G(Ci) feasible for an arbitrary green path of Ci by Feasible-Draw;

Let G’’ be a plane graph derived from G’ by contracting each for each G(Ci), 1 ≤ i ≤ l, to a single vertex vi;

Find a rectangular drawing D(G’’) of G’’ by Rectangular-Draw; Patch the drawings D(G(C1, D(G(C2 )) , …, D(G(Cl )) into D(G’’) to get an orthogonal

drawing D(G’) of G’. Obtain an orthogonal drawing D(G) of G by replacing the four dummy vertices a, b,

c and d in D(G’) with bendsend.

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Linear Algorithm Without Bad Cycles

Algorithm Orthogonal-Draw(G)begin Add four dummy vertices of degree two on four distinct outer edges as

the corners; Let G’ be the resulting graph; Obtain an orthogonal drawing D(G) of G by replacing the four dummy

vertices a, b, c and d in D(G’) with bendsend.

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Example Start:

Step 2:

Step 1:

Step 3:

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Linear Algorithm With Bad Cycles, same Algorithm

Start:

Step 2:

Step 1:

Step 3:

Doesn’t lead to orthogonal drawing due to bad cycle

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Bad Cycles in G’

G is 3-connected and thus has no 2-legged cycles

G’ has four 2-legged cycles, but none of those are bad cycles

Every bad cycle in G’ is a 3-legged cycle containing no corner

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Genealogical Tree

Tree of cycles in G, TC

In algorithm: tree of 3-legged bad cycles

TC can be found in linear time

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Example 3-legged cycles in

C and genealogical tree TC

C2, C3, C4 and C5 are the maximal bad cycles in C

C7 is the maximal bad cycle in C4

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Assignment & Labeling For each 3-legged cycle C in graph G

we define the following terms in a recursive manner, based on a genealogical tree. C is divided into three contour paths

P1, P2, and P3 by the three leg-vertices x, y and z

Each contour path is classified as either a green path or a red path

C is assigned an integer bc(C), the bend-count of C

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Three cases

1. C has no child-cycle, that is, l = 0

2. C has child-cycles C1, C2 , …, Cl , l ≥ 1, but none of them has a green path on C

3. Otherwise

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Case 1

C has no child-cycles: TC consists of a single vertex Classify all the three contour

paths of C as green paths Set bc(C) = 1

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Case 2

C has one or more child-cycles, none of them with a green path on C:

Classify all the three contour paths of C as green paths

Set bc(C) = 1 + the sum of the bend-counts of all child-cycles

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Case 3

Otherwise: Classify the contour paths in

which a child-cycle has a green path as green paths

Classify the other contour paths (if any) as red paths

Set bc(C) = the sum of the bend-counts of all child-cycles

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Properties

Each cycle has at least one green contour path

Assignment & Labeling can be done in linear time

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Feasible-Draw

Find an orthogonal drawing G(C) for a 3-legged cycle C

Feasible if the drawing has the following properties: Drawing has bc(C) bends At least one bend appears on the green

path The drawing is bound by six open

halflines, two at each point x, y and z

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Case 1C has no child-cycles: bc(C) = 1 Insert a dummy vertex t of degree two on an arbitrary

edge on the green path in C Now there are four corners:

Dummy vertex t Vertices x, y and z

No bad cycles, algorithm Rectangular-Draw can be used to find a rectangular drawing

After replacing t with a bend, an orthogonal drawing is created

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Case 2C has one or more child-cycles, none of them with a green path on C: Recursion: first find an orthogonal drawing for each child-cycle Construct a plane graph J from G(C) by contracting each child-cycle

to a single vertex Add a dummy vertex t Now, with Rectangular-Draw and removal of t, an orthogonal

drawing can be found Replace the contracted vertices with the original child-cycles. This

is always possible, there are 12 distinct embeddings, depeding on the three legs x, y and z and the chosen green path

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Case 3Otherwise Construct a plane graph J from G(C) by contracting each child-cycle

to a single vertex Replace one of the contracted cycles with a quadrangle (x1, t, y1,

z1) where t is a dummy vertext of degree two Recursion: find an orthogonal drawing for each child-cycle Now, with Rectangular-Draw and removal of t, an orthogonal

drawing can be found Replace the contracted vertices with the original child-cycles and

the quadrangle with its corresponding cycle.

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Linear AlgorithmAlgorithm Orthogonal-Draw(G)begin Add four dummy vertices of degree two on four distinct outer edges as the corners; Let G’ be the resulting graph; Let C1, C2 , …, Cl, be the maximal bad cycles in G’;

for each i, 1 ≤ i ≤ l, construct a genealogical tree TCi and determine green paths

and red paths for every cycle in TCi;

for each i, 1 ≤ i ≤ l, find an orthogonal drawing D(G(Ci)) of G(Ci) feasible for an arbitrary green path of Ci by Feasible-Draw;

Let G’’ be a plane graph derived from G’ by contracting each for each G(Ci), 1 ≤ i ≤ l, to a single vertex vi;

Find a rectangular drawing D(G’’) of G’’ by Rectangular-Draw; Patch the drawings D(G(C1, D(G(C2 )) , …, D(G(Cl )) into D(G’’) to get an orthogonal

drawing D(G’) of G’. Obtain an orthogonal drawing D(G) of G by replacing the four dummy vertices a, b,

c and d in D(G’) with bendsend.

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Orthogonal Grid Drawing An orthogonal drawing is called an

orthogonal grid drawing if all vertices and bends are located on integer grid points.

Dg D’g

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Compact

Dg is compact: There is at least one vertical line

segment of x-coordinate i for each integer i, 0 ≤ i ≤ W(W is the width of the grid)

There is at least one horizontal line segment of y-coordinate j for each integer j, 0 ≤ j ≤ H(H is the height of the grid)

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Bend-count and grid size relation

Let Dg be a compact orthogonal grid drawing of a plane graph G with b bends. Then W + H ≤ b + 2 * |E| – |V| – 2

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Orthogonal Drawings without Bends

Plane graphs with Δ ≤ 3 may have an orthogonal drawing without bends

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Conditions

Assume that G is a plane 2-connected graph with Δ ≤ 3 and four or more outer vertices of degree two.G has an orthogonal drawing without bends if and only if Every 2-legged cycle contains at least

two vertices of degree two Every 3-legged cycle contains at least

one vertex of degree two

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Example 2-legged cycles:

3-legged cycles: