Separator Theorems for Planar Graphs

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Separator Theoremsfor Planar Graphs

• Presented by
• Shira Zucker

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What is a Separator Theorem?

• S a class of graphs.
• An f(n)-separator theorem for S is a theorem of
  the following form
• There exist constants alt1, ßgt0, such that, if G
  is any n-vertex graph in S, the vertices of G can
  be partitioned into three sets A, B, C, such that
  no edge joins a vertex in A with a vertex in B,
  neither A nor B contains more than an vertices,
  and C contains no more than ßf(n) vertices.

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Theorem 1 (Lipton Tarjan)

• Let G be any n-vertex planar graph.
• The vertices of G can be partitioned into three
  sets A, B, C such that
• No edge joins a vertex in A with a vertex in B.
• Neither A nor B contains more than 2n/3 vertices.
• C contains no more than vertices.

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Facts

• Any n-vertex binary tree can be separated into
  two subtrees, each with no more than 2n/3
  vertices, by removing a single edge.
• Any n-vertex tree can be divided into two parts,
  each with no more than 2n/3 vertices, by removing
  a single vertex.
• A -separator theorem holds for the class of
  grid graphs.

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Most Graphs do not have a good Separator Theorem

• Theorem for every egt0, there exists a positive
  constant cc(e) such that almost all graphs G
  with n(2e)k vertices and ck edges have the
  property that, after the omission of any k
  vertices, a connected component of at least k
  vertices remains.
• Sparsity is not enough, but planarity is!!

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Some facts about planarity

• Let C be any simple closed curve in the plane.
  Removal of C divides the plane into exactly two
  connected regions, the inside and the outside
  of C.
• Any n-vertex planar graph with n3 contains no
  more than 3n-6 edges.
• A graph is planar iff it contains neither a
  complete graph on 5 vertices nor a complete
  bipartite graph on two sets of 3 vertices as a
  generalized subgraph.
• 4. Any planar graph can be triangulated.

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Lemma 1

• Let G be any planar graph. Shrinking any edge of
  G to a single vertex, we preserve planarity.
• Proof

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Corollary 1

• Let G be any planar graph. Shrinking any
  connected subgraph of G to a single vertex
  preserves planarity.
• Proof Immediate from Lemma 1 by induction on the
  number of vertices in the subgraph to be shrunk.

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Lemma 2

• Let G be any planar graph with
• Nonnegative vertex costs summing to at most 1.
• A spanning tree of radius r.
• Then the vertices of G can be partitioned
  into three sets A, B, C, such that
• No edge joins a vertex in A with a vertex in B.
• Neither A nor B has total cost exceeding 2/3.
• C contains no more than 2r1 vertices, one the
  root of the tree.

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Lemma 2 Proof
Assume no vertex has cost exceeding 1/3.
Otherwise, this vertex is C and the rest of the
graph is B. Make each face a triangle by adding
additional edges. Any nontree edge forms a simple
cycle with some of the tree edges. This cycle is
of length at most 2r1 or 2r-1. The cycle divides
the plane (and the graph) into two parts.
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Lemma 2 Proof Cont.

• Claim at least one cycle separates the graph so
  that neither the inside nor the outside contains
  vertices whose total cost exceeds 2/3.
• Proof Let (x,z) be the nontree edge whose cycle
  minimizes the maximum cost either inside or
  outside the cycle.
• Break ties by choosing the nontree edge whose
  cycle has the smallest number of faces on the
  same side as the maximum cost.

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Lemma 2 Proof Cont.

• Suppose without l.o.g that the graph is embedded
  so that the cost inside the (x,z) cycle is the
  cost outside the cycle.
• If the vertices inside the cycle have total cost
  2/3, the claim is true.
• Suppose the vertices inside the cycle have total
  cost gt 2/3.
• We show by case analysis that this contradicts
  the choice of (x,z).

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Consider the face which has (x,z) as a boundary
edge and lies inside the cycle.
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Lemma 3 - Definitions

• Let G be any n-vertex connected planar graph
  having nonnegative vertex costs summing to no
  more than 1.
• The vertices of G are partitioned into levels
  according to their distance from some vertex v.
• L(l) denotes the number of vertices at level l.
• r - the maximum distance of any vertex from v.
• Let r1 be an additional level containing no
  vertices.

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Lemma 3

• Given any two levels l1 and l2 such that
• Levels 0 through l1-1 have total cost not
  exceeding 2/3
• Levels l21 through r1 have total cost not
  exceeding 2/3.
• It is possible to find a partition A, B, C of
  the vertices of G such that
• No edge joins a vertex in A with a vertex in B.
• Neither A nor B has total cost exceeding 2/3.
• C contains no more than L(l1)L(l2)max0,2(l2-l1-
  1) vertices.

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Lemma 3 - Proof

• If l1l2,
• let A be all vertices on levels 0 through
  l1-1,
• B all vertices on levels l11 through r,
• and C all vertices on level l1.
• Then the Lemma is true.

0
l2
A
l1-1
l1
C
l11
B
r
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Proof Cont.
lt2/3

• Suppose l1ltl2. Delete the vertices at levels l1
  and l2 from G. This separates the remaining
  vertices of G into three parts vertices at
  levels 0 (l1-1), vertices at levels
• (l11) (l2-1) and vertices at levels l21
  and above.
• The only part which may have a cost exceeding
  2/3 is the middle part.

l1-1
l1
l11
l2-1
l2
l21
lt2/3
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Proof cont.

• If the middle part does not have cost exceeding
  2/3,
• let A be the most costly part of the three,
• let B be consist of the remaining two parts,
• and let C be the set of vertices at levels l1
  and l2.
• Then the lemma is true.

l1-1
l1
l11
C
l2-1
l2
l21
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Proof cont.

• Suppose the middle part has cost exceeding 2/3.
• Delete all vertices at levels l2 and above,
  and shrink all vertices at levels l1 and below to
  a single vertex of cost zero.
• These operations preserve planarity by Cor.
  1.
• The new graph has a spanning tree of radius
  l2-l1-1 whose root corresponds to vertices at
  levels l1 and below in the original graph.

l1-1
0
l1
l11
l2-1
l2
l21
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Proof Cont.

• Apply Lemma 2 to the new graph.
• Let A, B, C be the resulting vertex
  partition.
• Let A be the set among A and B having
  greater cost.
• Let C consist of the vertices at levels l1 and
  l2 in the original graph and the vertices in C,
  excluding the root of the tree, and let B contain
  the remaining vertices in G.
• By Lemma 2, A has total cost not exceeding 2/3.
  But AUC has total cost at least 1/3, so B also
  has total cost at most 2/3. Furthermore, C
  contains no more than L(l1)L(l2)2(l2-l1-1)
  vertices.

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Theorem 4

• Let G be any n-vertex planar graph having
  nonnegative vertex costs summing to no more than
  1.
• Then the vertices of G can be partitioned into
  three sets A, B, C such that
• No edge joins a vertex in A with a vertex in B,
• Neither A nor B has total cost exceeding 2/3,
• C contains no more than vertices.

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Theorem 4 Proof For a connected G
0

• Partition the vertices into levels according to
  their distance from some vertex v. Let L(l) be
  the number of vertices on level l. If r is the
  maximum distance of any vertex from v, define
  additional levels -1 and r1 containing no
  vertices.
• Let l1 be the level such that the sum of costs in
  levels 0 through l1-1 is less than ½, but the sum
  of costs in levels 0 through l1 is at least ½.
• (If no such l1 exists, then BCø satisfies the
  theorem.)

l0
k
l1-1
lt½
l1
gt½
l2
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Proof Cont.
0

• Let k be the number of vertices at levels 0
  through l1. Find a level l0 such that l0l1 and
  L(l0)2(l1-l0)2vk. Find a level l2 such that
  l11l2 and L(l2)2(l2-l1-1)2v(n-k).
• If two such levels exist, then by Lemma 3 the
  vertices of G can be partitioned into three sets
  A, B, C such that no edge joins a vertex in A
  with a vertex in B, neither A nor B has cost
  exceeding 2/3, and C contains no more than
  2(vkv(n-k)) vertices.

l0
k
l1-1
lt½
l1
gt½
l2
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Proof Cont.

• But,
• Thus, the theorem holds if suitable levels l0 and
  l2 exist.
• Suppose a suitable level l0 does not exist. Then,
  for il1,
• . Since L(0)1,
  this means
• , and
  .
• Thus, and
• which is a contradiction.

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Proof For a disconnected G

• Let G1, G2,Gk be the connected components of G,
  with vertex sets V1, V2,,Vk.
• If some connected component Gi has total vertex
  cost between 1/3 and 2/3, let AVi, BG\Vi and
  Cø.

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Proof For a disconnected G

• 2. If no connected component has cost gt1/3, let
  i be the minimum index such that the total cost
  of V1UV2UUVi exceeds 1/3.
• Let A V1UV2UUVi, let B Vi1UVi2UUVk, and let
  Cø.
• Since i is minimal and the cost of vi 1/3, the
  cost of A 2/3.

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Proof For a not connected G

• If some connected component (say Gi) has total
  cost gt 2/3, apply the theorem on Gi.
• Let A, B, C be the resulting partition.
• Let A be the set among A and B with greater
  cost.
• CC.
• Let B be the remaining vertices of G.

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An Algorithm for finding a good partition

• A representation of a planar embedding of a
  graph
• A list structure for the edges of the
  graph, its endpoints and 4 pointers, designating
  the edges immediately clockwise and
  counter-clockwise around each of the endpoints of
  the edge.
• Stored with each vertex is some incident
  edge.

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Example of a representation
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The algorithm

1. Find a planar embedding of G and construct a
   representation as described before.
2. Find the connected components of G and determine
   the cost of each one. If none has cost gt2/3,
   construct the partition as described in the proof
   of the theorem.
3. Find a BFS tree of the most costly component.
   Compute the level of each vertex and compute L(l)
   in each level l.

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The algorithm Cont.

• Back to slides 22-23.
• Construct a new vertex x to represent all
  vertices on levels 0 through l0.
• Construct a Boolean table with one entry
  per vertex. Initialize to true the entry for each
  vertex on levels 0-l0 and initialize to false the
  entry for each vertex on levels l01 through l2-1.

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The algorithm Cont.

• The vertices on levels 0-l0 correspond to a
  subtree of the BFS generated before.
• Scan the edges incident to this tree clockwise
  around the tree.
• When scanning an edge (v,w) with v in the tree,
  check the table entry for w. If it is true
  delete edge (v,w). If it is false change it to
  true, create (x,w) and delete (v,w).
• We get a planar representation of the shrunken
  graph, to which Lemma 2 is to be applied.

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Planar representation for the shrunken graph -
Example
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The algorithm Cont.

• Construct a BFS for the shrunken graph.
• Find an appropriate cycle, with cost 2/3 in the
  inside of the cycle.
• Use the found cycle and levels l0 and l2 (found
  in step 4) to construct a satisfactory vertex
  partition as described in the proof of Lemma 3.
  Extend this partition from the connected
  component chosen in step 2, to the entire graph
  as described in the proof of Theorem 4.
• ? All steps require O(n) time.

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Finding an appropriate cycle
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Application of the theorem

• Divide-and-conquer in combination with the
  theorem can be used to rapidly find good
  approximate solutions to certain NP-complete
  problems on planar graph.
• The maximum independent set problem.

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The maximum independent set problem.

• Theorem 5 Let G be an n-vertex planar graph with
  nonnegative vertex costs summing to no more than
  1, and let 0 e 1. Then there is some set C of
  O(v(n/e)) vertices whose removal leaves G with
  no connected component of cost exceeding e.
• C can be found in O(n log n) time.

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Proof for theorem 5

• If e 1/vn, let C contain all the vertices of G.
• Otherwise, apply the following algorithm
• Init Let Cø.
• Step Find some connected component K in G\C with
  cost gt e.
• Apply Theorem 4 to K, producing a partition
  A1, B1, C1 of its vertices.
• Let CCUC1.

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Proof Cont.

• Repeat the step until G\C has no component with
  cost gt e.
• Consider all components arising during the course
  of the algorithm. Assign a level to each
  component.

Level 0 Level 1 Level 2 Level 3
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Complexity

• Any two components at the same level are
  vertex-disjoint.
• Each level 1 component has cost gt e.
• For i1, each level i component has cost
  .
• The total cost of G is 1.
• Therefore, the total number of components of
  level i is at most .
• The maximum level k satisfy

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Complexity Cont.

• The time to split a component is linear in its
  number of vertices.
• Any two components at the same level are
  vertex-disjoint.
• ? The total running time of the algorithm is O(n
  log n).

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The size of the set C

• Let K1, K2, , Kl, of sizes n1, n2, , nl,
  respectively, be the components of some level
  i1.
• The number of vertices added to C by splitting
  K1, , Kl is at most .
• and
  .

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The size of C Cont.

• We have
• which leads to

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Algorithm for IS

• Apply Theorem 5 to G with ek(n)/n and each
  vertex having cost 1/n.
• Thus we find a set of vertices C of size
  O(n/vk(n)) whose removal leaves no connected
  component with more than k(n) vertices.
• 2. In each connected component of G\C, find a
  max IS by checking every subset of vertices. Form
  I as a union of max ISs, one from each component.

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Correctness

• Let I be a maximum IS of G.
• I\I cannot have vertices from G\C. (all
  possibilities were checked.)
• ? I\I can have vertices only from C.
• ? I-IO(n/vk(n)).
• G is planar ? G is 4-colorable ? I n/4.
• ?

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Correctness Cont.

• Thus, the relative error in the size of I ? 0 as
  n ? 8 if K(n) ? 8 as n ? 8.

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Complexity

• Step 1 of the algorithm requires O(n log n) by
  Theorem 5.
• Step 2 requires time on a
  connected component of ni vertices.
• The total time required by step 2

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Complexity Cont.

• Hence the entire algorithm requires
  
• 
  time.
• If we choose k(n)log n, we get an O(n²)-time
  algorithm with O(1/v(log n)) relative error.
• If we choose k(n)log log n, we get an O(n
  log n)-time algorithm with O(1/v(log log n))
  relative error.

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Better Results

• Alon, Seymour and Thomas gave a shorter proof to
  the theorem of Lipton Tarjan.
• They also proved a better theorem, with a smaller
  constant.

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The new Separator Theorem

• Let G be any n-vertex planar graph.
• The vertices of G can be partitioned into three
  sets A, B, C such that
• No edge joins a vertex in A with a vertex in B.
• Neither A nor B contains more than 2n/3 vertices.
• C contains no more than
  vertices.
• This theorem also holds for weighted
  graphs.