Divide and Conquer

Subject

• Series-Parallel Digraphs• Planarity testing

Series-Parallel Digraphs

Series-Parallel Digraphs

• Graph with a source and sink nodes• Source and sink are called poles• Defined recursively• Base case:

An edge with source and sink verticessource sink

Series composition of digraphs

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

Parallel composition of digraphs

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

Decomposition trees

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

Decomposition trees

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

Q

Decomposition trees

• G is a parallel composition of G1 and G2

where T1 and T2 are their parse treesso T has P-node root with children T1 and T2

P

T1 T2

Decomposition trees

• G is a series composition of G1 and G2

where T1 and T2 are their parse treesso T has S-node root with children T1 and T2

S

T1 T2

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

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

Closed component

Open component

Not a component

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

Δ-SP-Draw

• Upward embedding of SP-digraph• Drawing inside a bounding triangle Δ(Γ) which

is isosceles (שווה שוקיים) and right-handed• Series – one above the other• Parallel – one to the right of the other

Δ-SP-Draw invariants

1. Inside an isosceles right-angled triangle Δ(Γ)2. Top = Sink = t

Left = nothingBottom = Source = s

3. If (s, u) exist then between slopes –π/2 and –π/4there’s nothing but s

4. If (v, t) exist thenbetween slopes π/2 and π/4there’s nothing but t

t

s

v

u

Δ-SP-Draw

Δ-SP-Draw – base

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

G

t

s

Δ-SP-Draw – series

• If G is a series composition of G1 and G2• Recursively draw Γ1 and Γ2• Translate Γ2 so G2’s sink is on G1’s source• Extend bounding triangle Δ(Γ) accordingly

G1

G2

• If G is a parallel composition of G1 and G2• Recursively draw Γ1 and Γ2• Translate Γ2 to the “prescribed-region” of G2

(right of the dashes)• Extend bounding triangle Δ(Γ)

accordingly• Move sources and sinks to

upper and lower corners of Δ(Γ)

Δ-SP-Draw – parallel

G1

G2

“prescribed-region”

• (v, t) and (s, u) are rightmostedges on G1

• λv and λu are rays parallelto G1’s bounding triangle

• G2 will be translated toanywhere on the right of λv, λu and κ.

G1

G2v

u

κ

t

s

λv

λu

Lemma 3.3

• Let u and v be neighbors of the sourceso (s, u) is left of (s, v)

• Let λu and λv be rays of slope -π/4from u and v

• If invariant 3 holdsthen λu is below λv

s

u v

λu

λv

Lemma 3.3 – outcome

• If G2 is placed to the right of G1 and above λuwhere (s, u) is the rightmost edge of s on G1

• Then no vertex of G2 is in the wedge limited by the –π/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 proveds

u

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 Δ(Γ) is always empty,so G2’s left corner can beon the base of G1.

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)

Calculating distances

We’ll describe Γ by:b = length of the baseb‘ = vertical distance to λub‘’ = vertical distance to λv

v

u

κ

t

s

λv

λu

b’

b’’

b

Parallel distances

To describe parallel combinations we’ll use:Δx = horizontal distance from

Δ(Γ1) to Δ(Γ2)Δy = vertical distance from

bottom ray of Δ(Γ1) toleft corner of Δ(Γ2)

Δx

Δy

Δ-SP-labelinput : decomposition tree T of Goutput : labeling b, b’, b’’ for each subtree of Tif root = Q-node then

b(T) = b’(T) = b’’(T) = 2else

Δ-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 thenb(T) = b(T1) + b(T2)+2Δxif T2 is Q-node (transitive edge) then

b’(T) = b’’(T) = b(T)else

b’’(T) = b(T1) + 2Δx – Δy + b’’(T2)b’(T) = b’(T2) + Δy

Δ-SP-label

• Δx can be 0• Δy 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

Planarity Testing

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

Partitioning into Pieces

• Let a biconnected graph G contain a cycle C• Classify each edge of G not on C:

if 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.

Partitioning into Pieces

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

Separating cycle

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

separatingnonseparating

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 separating

cycle C’ consisting of subpath of C anda path of P

Lemma 3.4 proof• Let u and v be 2 attachments of P that are

consecutive in the circular ordering.• Let γ be a subpath of C between u and v

without any attachments.• Since P is connected there is a

path π in P between u and v.• Let C’ be the cycle obtained from C

by replacing γ with π.• Now γ is a piece on G with respect to C’• Let e be an edge on P not π.• e exist Because P is not a path.• So there is a piece of C’ other

than γ which contains e.• Thus C’ is a separating cycle in G.

C’

C

γ

π

e

Interlacement

• Each piece can be drawn either entirelyinside or outside of the cycle

• Interlacing pieces are ones that can’t be drawn on the same side of C without crossing

Interlacing pieces

Interlacement graph

• Vertices are pieces on G with cycle C• Edges are pairs of pieces that interlace – can’t

reside on the same side of C

CP1

P3P2

P4P1 P3

P2P4

Interlacement to planarity

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

CP1

P3

P2

P4

P1P3

P2P4

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

Planarity testing

0. Count edges and check Euler's formula1.Find pieces of G2.For each piece P that is not a path

test planarity by recursion3.Compute interlacement graph of the pieces4.Test if the interlacement graph is bipartite

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

Finding pieces

• Finding the pieces of G with respect to C by computing the connected components ofG without C.

• Takes O(n)

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 replacing

part of C with a path in P• Since P’ is biconnected

and C’ is separating– test planarity recursively

• Takes O(E(P’))

P2

Computing interlacement graph

For each 2 pieces P, Q• P, Q’s attachments are numbered along C• If attachments are not continuous in region

for each piece then P, Q are interlaced

1

3

2

1

3

2

4

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 it’s bipartite takes O(n2)

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)