Flow and Orthogonal

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Chapter 5
Flow and Orthogonal Drawings
Nimrod Milo and Ilan Orlov
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Final Example
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Goal
Efficiently drawing of orthogonal graphs
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Quality of a draw
Minimal number of bends
9 bends
6 bends
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Every edge considered as two directed edges !!!
Notations
G Orthogonal graph
G Original graph
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Lemma 1
In a planner orthogonal drawing, the sum of the
measures around a vertex is equal to 2?
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Lemma 2.1
For each internal face of a planar orthogonal
drawing. The sum of the measures of the
vertex-angels and bend-angles inside f is equal
to ?(p-2) where p is the number of such angels.
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Lemma 2.2
For each external face of a planar orthogonal
drawing. The sum of the measures of the
vertex-angels and bend-angles inside f is equal
to ?(p2) where p is the number of such angels
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Notations
a(f) number of vertex-angles in f
Counterclockwise edge with respect to face f
f - face
D(f) - Counterclockwise edge inside the face f
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Notations
a(u?v) ??/2 The angle at vertex u formed by edge
(u?v) and the next counterclockwise around u
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Notations
ß(u?v) number of bends along (u?v) With the
?/2 angle on the left hand side.
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Notations
a ?
ß ?
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a
1
4
ß
0
4
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Flow Network Model
A flow network N for a digraph G(V,E) defined
such that

• Each source v has a production s(v).
• Each sink u has a consume rate s(u).
• Total production total consumption

• For each edge (u,v) in E
• ?(u,v) a lower bound of flow
• µ(u,v) - Capacity of flow
• ?(u,v) Cost per unit of flow

• A flow ? (? IN ? IN) in Network N defined such
  that
• for each (u,v) in E
• ?(u,v) ?(u,v) µ(u,v)
• The Cost of flow ? ??(u,v) ?(u,v)
• The objective is a Minimum cost of flow

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Building network from graph

• One unit of flow ?/2
• The nodes of N are the vertices and faces of G

• A vertex-node v in V(N) produced flow s(v)4

• A face-node f in V(N) consumes flow
• s (f) 2a (f) 4 if f is internal face
• s (f) 2a (f) 4 if f is external face

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Building network from graph

• For each directed edge (u,v) in E with faces f,
  g on its left and right respectively we define
  to edges in E(N)

• For edge (u,f) ?(u,f) 1, µ(u,f) 4 and
  ?(u,f)0 (figure a)

• For edge (f,g) ?(f,g) 0, µ(f,g) 8 and
  ?(f,g )1 (figure b)

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Building network from graph (Contd.)
Lemma 3 Let G an embedded planar graph with n
vertices. The flow Network N built from G has
O(n) nodes and edges, And can be built in O(n)
time.
Theorem 1 Let G an embedded planar graph and N
be the flow Network built from G. for the edge
(u,v), let (u,f) and (f,g) be the associated
edges in N

• ?(u,f) a(u,v)

• ?(f,g) ß(u,v)

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Orthogonalize (algorithm 5.1)
Input planar Graph G with n vertices (Maximum
degree of 4)
Output orthogonal represntation of G with
minimum number of bends
Time Complexity O(n²log n) Space Complexity
O(n)

• The algorithm
• Construct the flow Network N from G
• Compute Flow of minimum cost for N.
• Compute the Orthogonal representation of G
  associated with the minimum cost flow.

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Compacting of Orthogonal Representation

• Goal
• Assign integer length to the segments
• No crossing, no overlaps
• Minimum area, fast algorithm
• our case
• area O((nb)2) b no. bends
• time O((nb))

First, assume rectangular faces !!
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Rectangular faces

• Each internal face has the shape of a rectangle
• in the orthogonal representation H
• H has at most 4 bends, placed at the corner of
  the external face.
• Edge of G edge of the H, excepts possibly for
  eight segments on the external face

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Tidy-Rectangle-Compact algorithm
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For each edge set Minimum flow 1 Cost 1
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Run algorithm for finding flow with minimal cost
flow 1
flow 2
flow 3
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Setting edge length Flow on the
corresponding edge in the flow network
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For each edge set Minimum flow 1 Cost 1
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flow 1
flow 2
flow 3
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Setting edge length Flow on the
corresponding edge in the flow network
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Combining the new vertical and horizontal lengths
Running time O(n7/4 logn)
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Running time O(n7/4 logn)
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Different compaction algorithm

• Linear time
• Minimizes the height, width, area of the drawing
• Does not guarantee the minimum total edge length

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Different compaction algorithm
1. Replacing bends with fictitious vertices
2. Orienting the horizontal edges from left to
right
3. Contracting maximal paths of vertical edges to
a vertex
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compaction algorithm based on s-t-graph
compaction algorithm
We get a planer s-t-graph.
Denote it as Hhor.
Using same process we get Hver.
Assign unit weights to the edges to both
s-t-graphs edges
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Compute optimal weighted topological numbering X
and Y As in previous lesson
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Running time O(n)
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General orthogonal representations

1. Add an isolated vertex

2. Insert a vertex along an edge
H
3. Add an edge
H
G ? H ? H ? Hcompact? Hcompact_without_extra
edges/vertices
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Final Theorem 5.4
Given an embedded planar graph G with n vertices
of degree at most 4, and an orthogonal
representation H of G with b bends, a planar
orthogonal drawing of G with integer coordinates
and area O( (nb)2 ) can be constructed in O(nb)
time.
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Optimal-Orthogonal (algorithm 5.4)
Input planar Graph G with n vertices (Maximum
degree of 4)
Output planar orthogonal grid drawing D of G
with area O(n²) and minimum number of bends
Time Complexity O(n²log n) Space Complexity
O(n)

• The algorithm
• H ? Orthogononalize on graph G.
• Refine H into an orthogonal representation H
  with rectengular faces.
• D ? Fast-Orthogonal-Compact(H)
• D ? D while ignoring all the fictious edges and
  vertices.

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Constrains

• We can easily modify algorithm 5.4 to include two
  types of constrains

• Vertex-angle constrains by setting the lower /
  upper bound of a(u,v)

• Bend Constrains by setting the lower / upper
  bound of ß(u,v)

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Minimizing number of bends Locally
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Minimizing number of bends Locally
For a face-angle or a vertex angle we
Define p/2 inflex p - flat 3p/2 - reflex
For a given orthogonal representation G we define
a curve C as an Elementary Transformation If the
curve C intersect G only in inflex, flat or
reflex angles.
For each vertex traversed by Curve C the
transformation subtract p/2 from the angle C
enters, and adds p/2 to the angle where C exit.
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For a Curve C we define flat(C),inflex(C),reflex(C
) as the number of flat, inflex and reflex
angles in C respectively
?(C) flat(C) inflex(C) reflex(C)
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Theorem 5.6 an Orthogonal representation G has
the minimum Numbers of bend iff for every cycle
C in G ?(C) 0
The proof of Theorem 5.6 is based on looking in
the flow network Corresponding with G and flow
in the cycle C. the result is that the flow
cant be a negative number.
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Elementary Transformation
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Elementary Transformation
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Elementary Transformation
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Elementary Transformations example