# Orthogonal Drawing by nikeborome

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```									Orthogonal Drawing

Kees Visser
Overview
   Introduction
   Orthogonal representation
   Flow network
   Bend optimal drawing
Introduction
 Orthogonal drawing
 Given embedding
 Only vertical or horizontal line segments
 Maximum degree of 4
Introduction
 Uses in VLSI circuits
 Minimize bends
 Finding an orthogonal drawing of
graph G with the minimum number of
bends without a fixed embedding is
NP-Complete.
Introduction
 Tamassia and Garg and Tamassia
found algorithms to find a minimal
bend orthogonal drawing of graph G
in polynomial time.
Orthogonal Representation
 Planar representation
 Set of circularly ordered edge list P(F)
 One for each face
   P(F1) =   (e8,   e7,   e11, e6, e3)
   P(F2) =   (e1,   e2,   e10, e9, e11, e7, e8)
   P(F3) =   (e6,   e9,   e10, e5, e4)
   P(F0) =   (e1,   e2,   e5, e4, e3)
Orthogonal representation
 Definitions:
 An edge appearing in the same list twice
is a bridge
 An angle formed by to edges is a vertex-
angle
 An angle formed by two line segments at
a bend is a bend-angle
 Both vertex angles and bend angles are
90, 180, 270 or 360 degrees
Orthogonal representation
 Facts:
 The sum of vertex-angles around any
vertex is 360 degrees
 The sum or the angles inside any facial
polygon is (2p-4)90 degree, and the sum
of the angles of the outer facial polygon
is (2p+4)90 degrees, where p is the
number of edges
Orthogonal representation
 Orthogonal representation
 1 list for each face
 Each element in the list has 3 values, the
edge number (Er), a bit string and the
angle formed in the face from this edge
to the next edge (Ar).
 The bit string indicated the bends in the
line. A 0 indicates a 90 degree bend, 1
means 270 and ε means a straight line
Orthogonal representation
 R(F1) = ((e8, ε, 180),(e7, ε, 90),(e11, ε,
90),(e6, ε, 180),(e3, 0, 90))
 R(F2) = ((e1, 0, 90), (e2, 00, 90), (e10,
ε, 180), (e9, ε, 180), (e11, ε, 270), (e7,
ε, 180), (e8, ε, 90))
Orthogonal representation
 Necessary properties of a orthogonal
representation:
 (p1) There is some planer graph whose
planer representation is given by the e-
fields of the list in R
 (p2) For each pair of elements r and r’ in
R with Er=Er’, string Sr’ can be obtained
by applying bitwise negation to the
reversion of Sr’
Orthogonal representation
 (p3) For each element r in R, define the
rotation p(r) as follows: p(r) = |Sr|0 -
|Sr|1 + (2-Ar/90)
 (p4) For each vertex v, the sum of the
vertex angles around v given by the A-
fields in R is equal to 360 degrees
Flow network
 A flow network N is a directed graph
with two disjoint non empty sets of
sources and sinks
 Each arc e of N is labelled with 3
nonnegative integers
 A lower bound λ(e)
 A capacity μ(e)
 A cost c(e)
Flow network
 The flow Ф(e) must equal or higher
then the lower bound and lower of
equal then the capacity.
 σ(u) is the production of source u
 -σ(u) is the consumption of sink u
Bend-optimal drawing
 1 unit of flow represents an angle of 90
degrees
 Each vertex and face is a node
 Each vertex Node is a source (Uv), σ(Uv)
=4
 Each face Node is a sink -σ(UF), -σ(UF) =
2p(F)-4 (if F is an inner face), -σ(UF) =
2p(F)+4 (if F is the outer face). p(F) is
the number of vertex-angles in face F
Bend-optimal drawing
Bend-optimal drawing
 The arc set of network N consists of the vertex
arcs (Av) and face arcs (AF)
 The set of Av consists of arcs from every vertex
node to every face node the vertex borders.
 Flow Ф(Uv, UF) is the sum of vertex angles at
vertex v inside face F.
 λ(Uv, UF) is equal to the number of vertex-
angles at v inside face F.
 μ(Uv, UF) = 4
 c(Uv, UF) = 0
Bend-optimal drawing
Bend-optimal drawing
 The set of AF consists of arcs from every
face F that shares an edge with face F’
 Flow Ф(UF, UF’) is the number of bends
with an angle of 90 degrees inside face F
along the edges which are common to F
and F’
 λ(UF, UF’) = 0
 μ(UF, UF’) = +∞
 c(UF, UF) = 1
Bend-optimal drawing
Bend-optimal drawing
 Theorem: Let G be a plane graph, let N
be the network constructed from G. For
each integer flow Ф in network N, there
is an orthogonal representation R that
represents an orthogonal drawing D of G
and whose number of bends is equal to
the cost of the flow Ф. In particular, the
minimum cost flow can be used to
construct a bend-optimal orthogonal
drawing of G.
Bend-optimal drawing
 Proof
 We construct a orthogonal
representation R of G by computing the
a- and s-fields from the flow values in Av
and AF.
 L = |(R(v,F)| and R(v,F) ={r1,r2,…,rL},
where 1 =< L =< 4
 Ar1 = 90(Ф(Uv, UF)- L + 1) and Ari = 90
for each i, 2 =< i =< L
Bend-optimal drawing
   Sr1 = 0 Ф(UF, UF’)1 Ф(UF’, UF)
   Sr’1 = 0 Ф(UF’, UF)1 Ф(UF, UF’)
   Sri = Sr’i = ε for each i 2 =< i =< L
   Example:
 L = 3, Ф(UF, UF’) = 2 and Ф(UF’, UF) = 1
 Sr1 = 001 and Sr2 = Sr3 = ε
Bend-optimized drawing
 R(F) needs to satisfy Properties (p1)-(p4)
 (p1) There is some planer graph whose planer
representation is given by the e-fields of the list
in R
 (p2) For each pair of elements r and r’ in R with
Er=Er’, string Sr’ can be obtained by applying
bitwise negation to the reversion of Sr’
 (p3) For each element r in R, define the rotation
p(r) as follows: p(r) = |Sr|0 - |Sr|1 + (2-Ar/90)
 (p4) For each vertex v, the sum of the vertex
angles around v given by the A-fields in R is
equal to 360 degrees
Bend-optimized drawing
 (p1) is automatically satisfied since we
have build R from a planar representation
 (p2) easily follows from the assignment of
the S-values.
 (p4) follows from the definition that every
source produces 4 units.
 (p3) Solving the equation by replacing with
the flow equations makes this satisfied.
Look in the book for a more precise proof of
this.
Conclusion
 Garg and Tamassia have shown that
the minimum cost flow problem in
this specific network can be solved in
time O(n7/4√(log n))
Questions?

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