Divide and Conquer

1
Divide and Conquer
2
Subject

• Series-Parallel Digraphs
• Planarity testing

3
Series-Parallel Digraphs
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Series-Parallel Digraphs

• Graph with a source and sink nodes
• Source and sink are called poles
• Defined recursively
• Base caseAn edge with source and sink vertices

source
sink
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Series composition of digraphs

• Identifying the sink of one and the source of the
  other as one vertex

6
Parallel composition of digraphs

• Identifying as one vertex both their sources
• Identifying as one vertex both their sinks

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Decomposition trees

• Associating SP digraphs with Parse Trees
• Binary trees
• S-nodes, P-nodes and Q-nodes
• Defined recursively

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Decomposition trees

• G is a single edge so T is a single Q-node

Q
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Decomposition trees

• G is a parallel composition of G1 and G2where T1
  and T2 are their parse treesso T has P-node root
  with children T1 and T2

P
T1
T2
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Decomposition trees

• G is a series composition of G1 and G2where T1
  and T2 are their parse treesso T has S-node root
  with children T1 and T2

S
T1
T2
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Decomposition trees

• If a node has the same type as his parentit will
  be the right child
• T is unique for G
• If G has n nodes then T has O(n) nodes
• T can be created in O(n) time so will be assumed
  to be part of the input

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Components

• Let C be a maximal path same node type on T
• Let u1 uk children of C not on C
• Component is the part of G associated with a
  subsequence ui .. uj where 1 i j k
• Component is closed when C is of S-nodes
• Otherwise it is open

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Closed component
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Open component
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Not a component
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Right-pushed embedding

• (u, v) is transitive in G if there is another
  path from u to v
• Drawing transitive edge (u, v) on the right of
  all components with u and v poles
• O(n2) drawing area

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?-SP-Draw

• Upward embedding of SP-digraph
• Drawing inside a bounding triangle ?(G) which is
  isosceles (???? ??????) and right-handed
• Series one above the other
• Parallel one to the right of the other

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?-SP-Draw invariants

1. Inside an isosceles right-angled triangle ?(G)
2. Top Sink tLeft nothingBottom Source
   s
3. If (s, u) exist then between slopes p/2 and
   p/4theres nothing but s
4. If (v, t) exist thenbetween slopes p/2 and
   p/4theres nothing but t

t
v
u
s
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?-SP-Draw
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?-SP-Draw base

• If G is a single edge
• Draw it as a vertical edge with length 2
• Bounding triangle ?(G) will have width 1

t
G
s
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?-SP-Draw series

• If G is a series composition of G1 and G2
• Recursively draw G1 and G2
• Translate G2 so G2s sink is on G1s source
• Extend bounding triangle ?(G) accordingly

G1
G2
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?-SP-Draw parallel

• If G is a parallel composition of G1 and G2
• Recursively draw G1 and G2
• Translate G2 to the prescribed-region of
  G2(right of the dashes)
• Extend bounding triangle ?(G)accordingly
• Move sources and sinks toupper and lower corners
  of ?(G)

G2
G1
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prescribed-region

• (v, t) and (s, u) are rightmostedges on G1
• ?v and ?u are rays parallelto G1s bounding
  triangle
• G2 will be translated toanywhere on the right
  of ?v, ?u and ?.

?
?v
t
v
G2
u
G1
s
?u
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Lemma 3.3

• Let u and v be neighbors of the sourceso (s, u)
  is left of (s, v)
• Let ?u and ?v be rays of slope -p/4from u and v
• If invariant 3 holdsthen ?u is below ?v

u
v
?v
s
?u
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Lemma 3.3 outcome

• If G2 is placed to the right of G1 and above
  ?uwhere (s, u) is the rightmost edge of s on G1
• Then no vertex of G2 is in the wedge limited by
  the p/4 ray from the neighbors of s
• Hence if invariant 3 is held onG1 and G2 it is
  also held on G
• Symmetrically invariant 4 can be proved

u
s
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Therefore

• Proving Invariants 3 and 4 guarantee thats and t
  can be moved as described in the algorithm
  without creating crossings.
• Invariant 2 states that left cornerof ?(G) is
  always empty,so G2s left corner can beon the
  base of G1.

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Area of the triangle

• So the base of resulting triangle is equals the
  sum of bases of both triangles.
• For a graph with n verticesthe length of the
  base is 2n.
• So the area needed is O(n2)

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Calculating distances

• Well describe G by
• b length of the base
• b vertical distance to ?u
• b vertical distance to ?v

?
b
?v
t
v
b
u
s
?u
b
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Parallel distances

• To describe parallel combinations well use
• ?x horizontal distance from?(G1) to ?(G2)
• ?y vertical distance frombottom ray of ?(G1)
  toleft corner of ?(G2)

?x
?y
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?-SP-label

• input decomposition tree T of G
• output labeling b, b, b for each subtree of
  T
• if root Q-node then
• b(T) b(T) b(T) 2
• else?-SP-label of T1 and T2 (the subtrees)
• if root S-node then
• b(T) b(T1) b(T2)
• b(T) b(T1)
• b(T) b(T2)
• else root P-node then
• b(T) b(T1) b(T2)2?x
• if T2 is Q-node (transitive edge) then
• b(T) b(T) b(T)
• else
• b(T) b(T1) 2?x ?y b(T2)
• b(T) b(T2) ?y

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?-SP-label

• ?x can be 0
• ?y can be b(T1)
• The area needed is O(n2)
• The algorithm runs in O(n)and needs O(n) space
• end of Series-Parallel Digraphs

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Planarity Testing
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Planarity Testing

• Graph is planar if E 3V - 6 (Euler formula)
• Graph is planar iff all its connected components
  are planar
• Graph is planar iff all its biconnected
  (connected by 2 edges per vertex) components are
  planar
• Preprocessing into connected and biconnected
  components the problem is restricted to
  biconnected graphs

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Partitioning into Pieces

• Let a biconnected graph G contain a cycle C
• Classify each edge of G not on Cif 2 edges have
  a path between them with no vertex of C, they
  have the same class.
• Subgraph induced by edges in a class is called a
  piece of G.

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Partitioning into Pieces
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Attachments

• Vertices of piece P which are also on cycle C are
  called attachments
• Since G is biconnected each piece has at least 2
  attachments
• Cycle C induce a circular ordering on the
  attachments of P

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Separating cycle

• Cycle C is separating if it has at least 2 pieces
• If is has one or no pieces it is nonseparating

separating
nonseparating
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Lemma 3.4

• Let G be a biconnected graph
• Let C be a nonseparating cycle of G
• Let P be a piece on C
• If P is not a path then G has a separatingcycle
  C consisting of subpath of C anda path of P

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Lemma 3.4 proof

• Let u and v be 2 attachments of P that are
  consecutive in the circular ordering.
• Let ? be a subpath of C between u and vwithout
  any attachments.
• Since P is connected there is a path p in P
  between u and v.
• Let C be the cycle obtained from Cby replacing
  ? with p.
• Now ? is a piece on G with respect to C
• Let e be an edge on P not p.
• e exist Because P is not a path.
• So there is a piece of C other than ? which
  contains e.
• Thus C is a separating cycle in G.

p
?
e
C
C
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Interlacement

• Each piece can be drawn either entirelyinside or
  outside of the cycle
• Interlacing pieces are ones that cant be drawn
  on the same side of C without crossing

Interlacing pieces
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Interlacement graph

• Vertices are pieces on G with cycle C
• Edges are pairs of pieces that interlace cant
  reside on the same side of C

P3
P2
P1
C
P1
P3
P4
P2
P4
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Interlacement to planarity

• If G is planar graph then its interlacement graph
  must be bipartite

P3
P2
P1
P1
C
P3
P4
P2
P4
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Theorem 3.8

• A biconnected graph G with cycle C is planar iff
• For each piece P, adding P to C is planar
• The interlacement graph of pieces of G with cycle
  C is bipartite

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Planarity testing

• 0. Count edges and check Euler's formula
• Find pieces of G
• For each piece P that is not a pathtest
  planarity by recursion
• Compute interlacement graph of the pieces
• Test if the interlacement graph is bipartite

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Preliminaries

• The algorithm receives a biconnected graph G
• G has n vertices and at most 3n-6 edges
• Also received is a separating cycle C
• Returns whether G is planar or not

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Finding pieces

• Finding the pieces of G with respect to C by
  computing the connected components ofG without
  C.
• Takes O(n)

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Testing piece P planarity

• For each piece P that is not a path
• Let P be graph of adding P to C
• Let C be cycle in P by replacingpart of C with
  a path in P
• Since P is biconnectedand C is separating
  test planarity recursively
• Takes O(E(P))

P2
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Computing interlacement graph

• For each 2 pieces P, Q
• P, Qs attachments are numbered along C
• If attachments are not continuous in region for
  each piece then P, Q are interlaced

1
1
4
2
2
3
3
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Computing interlacement graph

• Checking one piece vs. all others takes O(n)
• All vs. all takes O(n2)
• Interlacement graph has O(n) vertices and O(n2)
  edges
• Checking if its bipartite takes O(n2)

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Overall runtime

• Each recursive invocation takes O(n2)
• Each recursion is associated with at least one
  edge of G
• There are O(n) edges
• Runtime is O(n3)