External Memory Multi-Way Planar Graph Separation

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External Memory Multi-Way Planar Graph Separation

• (Arge, Brodal, Toma)

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The Graph Separation Problem

• Given an undirected graph G (V ,E) and a
  function f N ?N, we would like to find a subset
  S ? V of size f(V) such that the removal of S
  disconnects G into two subgraphs, each of size at
  most 2V/3.

vV separator vertices
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Planar Graphs

• A planar graph is a graph that can be embedded to
  the plane so that no two edges cross except at
  the endpoints.
• Kuratowski's theorem
• a finite graph is planar ? it does not contain a
  subgraph that is homeomorphic to K5 or K3,3.

nonplanar
planar
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The Dual of a Planar Graph

• Given a planar graph G, its dual G is defined as
  follows
• Place a vertex in each face of G (including the
  outer face).
• For any two faces fi and fj adjacent to a common
  edge e (u, v) of G, add an edge e (fi, fj)
  in G.

G can be computed from G using O(sort(N)) I/Os.
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Dual Spanning Tree

• Lemma
• T is a spanning tree of G ? T(E\T) is a
  spanning tree of G

E\T T T
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Multi-Way Planar Graph Separation

• Lipton and Tarjan proved that any planar graph
  has an O(vV )-separator.
• Hutchinson et al. showed how it can be done using
  O(sort(V)) I/Os, assuming that a BFS tree of the
  graph is given.
• For any parameter R, this result can be used
  recursively to partition a planar graph into
  O(V/R) subgraphs with O(R) vertices each, using
  O(V/vR) separator vertices.
• This yields an O(log(V/R)sort(V)) I/Os
  algorithm. We will see an O(sort(V)) algorithm
  (assuming a BFS tree is given).

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Multi-Way Planar Graph Separation (2)

• Given a graph G we will assume that the following
  holds
• G is connected.
• A planar embedding of G is given.
• A BFS tree T of G is given, and is represented
  implicitly by storing with each vertex in G its
  parent in T and each edge of G is marked as being
  either a tree or a non-tree edge.
• W.l.o.g, we also assume that G is triangulated.
  If this is not the case, it can be triangulated
  using O(sort(N)) I/Os (Hutchinson et al.), and
  the added edges can be marked and removed in the
  final phase.

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Partitioning a Tree(Gazit et al.)
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• For a vertex v, w(v) is the number of vertices in
  the subtree rooted at v.
• v is called R-critical if v is not a leaf and
• ?w(v)/R? gt ?w(v)/R? for all children v of v.

• e and e are equivalent if there is a path
  connecting them that avoids R-critical vertices.
• The graphs induced by the equivalence classes are
  called the R-bridges of T.
• The attachment vertices of a R-bridge are the
  vertices in the bridge that are also R-critical.

• Combinatorial Results
• T has at most 2N/R-1 R-critical vertices.
• If T has degree d, then it has d(2N/R - 1)
  R-bridges.
• An R-bridge contains at most R1 vertices.
• An R-bridge has at most 2 attachment vertices.

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Case 1 Planar graph with bounded height spanning
tree

• G is triangulated, and thus T (E\T) is a
  binary tree.
• Conclusion
• T has at most 4V/R R-bridges (each of size at
  most R1), and thus is has a total of at most
  8V/R attachment vertices.
• If those vertices are removed, T breaks into
  O(V/R) subtrees (the R-bridges) of size O(R).

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Case 1 Planar graph with bounded height spanning
tree (2)

• Any edge in T has a dual edge in (E\T). Adding
  that edge to T creates a cycle of size O(H),
  where H is the height of T.
• The R-bridge B has (at most) 2 attachment
  vertices. The (at most) 2 edges in B incident to
  those vertices define 2 cycles in G.
• The faces of G that are inside one of these
  cycles but outside the other are exactly the
  faces corresponding to the vertices in B.
• B contains at most R1 vertices, and thus the 2
  edges define a subgraph of G of size at most
  3(R1).

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Case 1 Planar graph with bounded height spanning
tree (3)
Edges of G Cycle Edges Edges of T R-critial
vertices
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Case 1 Planar graph with bounded height spanning
tree (4)

• Bottom line
• There are O(N/R) R-bridges in T.
• Each defines at most 2 cycles in T.
• Each cycle is of size O(H) vertices.
• Thus G is partitioned to O(N/R) subgraphs of size
  O(R) using O((N/R)H) separator vertices.

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Case 1 I/O Efficiency

• Computing G and T can be done in O(sort(N))
  I/Os.
• BFS and DFS traversals on undirected trees are
  done with O(sort(N)) I/Os, and thus
• Calculating the weights is also done using
  O(sort(N)) I/Os.
• Once we have the weights, the R-bridges are
  computed with O(sort(N)) I/Os using a simple tree
  traversal.

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Case 1 I/O Efficiency (2)

• To compute the separator vertices and subgraphs
• Scan through the R-bridges and for each vertex
  output the 3 vertices defining the dual face to a
  list L, along with a unique identifier of the
  R-bridge.
• Sort L by vertex and then by the identifier, and
  remove vertices with more than one identifier.
  Those are the separator veritces.
• Remove duplicate vertices.
• This is done using O(sort(N)) I/Os.
• Total cost O(sort(N)) I/Os.

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Case 2 General planar graph

• Let L(i) be the total number of vertices on
  levels 0 through i of T.
• Given parameters X and Y lt N, define
• starter levels the levels i such that (L(i),
  L(i1) contains a multiple of ?N/X?.
• cutter levels the first level above and below
  some starter level, which contains at most Y
  vertices.
• There are at most X starter levels.
• Between each two consecutive starter levels there
  are at most ?N/X? vertices.

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Case 2 General planar graph (2)

• Partition G to O(X) subgraphs by grouping
  vertices between two consecutive cutter levels
  together.

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Case 2 General planar graph (3)

• If the two cutter levels defining a subgraph Gi
  are within two consecutive starter levels, then
  Gi has size O(N/X).
• Solve the problem recursively on Gi.
• Otherwise, each of the levels of T in Gi have
  more than Y vertices. Thus Gi has a spanning tree
  of height O(N/Y).
• Solve using the algorithm for Case 1.

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Case 2 General planar graph (4)

• The part of T that falls within a subgraph Gi is
  not a BFS tree for Gi, since it is not connected.
• In order to obtain a BFS tree for a subgraph,
  introduce a fake root vertex and connect it to
  all vertices just below the cutter level defining
  Gi.
• Assume that T is given level-by-level.
• Then the BFS trees for all subgraphs can be
  computed in O(scan(N)) I/Os.
• Each fake root replaces at least one vertex on
  the cutter level and thus the total size of the
  subgraphs remains O(N).
• Fake vertices and edges can be marked and removed
  later using another O(scan(N)) I/Os.

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Case 2 Analysis

• Choose Y N / vR.
• Bounded subgraph Gi of size Ni has height vR, and
  thus can be partitioned as in case 1 to O(Ni / R)
  subgraphs, each of size O(R), using O((Ni /R)vR)
  O(Ni /vR) separator vertices.
• In addition, the at most X cutter levels
  contribute another O(X Y) O(X N/vR)
  separator vertices.

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Case 2 Analysis (2)

• Thus, the total number of separator vertices is
  given by S(N) O(X N/vR) O(N/vR) X
  S(N/X) (and S(R) 0)
• Choose X (M/B2)1/4.
• For now, assume that RgtBvM.
• Thus XN/vR O(N/B).
• Therefore S(N) O(N/B)(M/B2)1/4
  S(N/(M/B2)1/4).
• This solves to O(N/B)log(M/B2)1/4(N/R), which is
  O(sort(N)) under the assumption that M gt B2e.

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Case 2 Analysis (3)

• Final result
• G (V, E) was partitioned into O(N/R)
  subgraphs, each of size O(R), using a set of
  O(sort(N)) separator vertices in O(sort(N)) I/Os.

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Case 2 I/O Efficiency

• Representing T level-by-level and computing the
  BFS level for each vertex can be done using
  O(sort(N)) I/Os using standard tree algorithms.
• Not including the I/Os needed to handle bounded
  height subgraphs, one recursion step can be
  performed in O(scan(N)) I/Os.
• Therefore, the recurrence is T(N) N/BXT(N/X),
  which is O(sort(N)).
• Bounded height subgraphs can be separated
  immediately, using a total of O(sort(N)) I/Os.

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Case 2 I/O Efficiency (2)

• So far we assumed that R gt BvM.
• If R BvM lt M, use the above algorithm to
  partition G into subgraphs of size O(M) and then
  load each subgraph into memory in turn and apply
  the algorithm of Lipton and Tarjan recursively
  until all subgraphs have size O(R).
• This only requires an extra O(N/B) I/Os and
  introduces O(M/vR) separator vertices in each of
  the O(N/M) subgraphs, for a total of O(N/vR)
  vertices.

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Summery

• Weve actually seen an O(sort(N)) reduction from
  the multi-way planar graph separation problem to
  planar BFS.
• Arge et al. also showed how the multi-way
  separation of a planar graph can be used to solve
  the SSSP problem in O(sort(V)) I/Os.
• In a different paper, Arge et al. showed that
  planar DFS can be reduced to planar BFS in
  O(sort(V)) I/Os.
• Since BFS can be trivially reduced to SSSP by
  assigning all edges weight 1, all fundamental
  problems on planar undirected graphs are
  equivalent.