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External Memory Multi-Way Planar Graph Separation

External Memory Multi-Way Planar Graph Separation. (Arge, Brodal, Toma). The Graph Separation Problem.

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External Memory Multi-Way Planar Graph Separation

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  1. External Memory Multi-Way Planar Graph Separation (Arge, Brodal, Toma)

  2. 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. √V separator vertices

  3. 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

  4. 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.

  5. Dual Spanning Tree Lemma: T is a spanning tree of G  T’=(E\T)* is a spanning tree of G* E\T T T’

  6. Multi-Way Planar Graph Separation • Lipton and Tarjan proved that any planar graph has an O(√V )-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/√R) 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).

  7. 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.

  8. Partitioning a Tree(Gazit et al.) 24 • 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 > w(v’)/R for all children v’ of v. 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 R+1 vertices. • An R-bridge has at most 2 attachment vertices. • 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. 23 11 11 10 10 9 5 4 8 4 2 1 1 5 2 3 4 1 2 1 2 1 1 R = 9

  9. 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/RR-bridges (each of size at most R+1), 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).

  10. 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 R+1 vertices, and thus the 2 edges define a subgraph of G of size at most 3(R+1).

  11. Case 1: Planar graph with bounded height spanning tree (3) Edges of G Cycle Edges Edges of T’ R-critial vertices

  12. 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.

  13. 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.

  14. 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.

  15. 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 < N, define: • starter levels: the levels i such that (L(i), L(i+1)] 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.

  16. Case 2: General planar graph (2) • Partition G to O(X) subgraphs by grouping vertices between two consecutive cutter levels together.

  17. 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.

  18. 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.

  19. Case 2: Analysis • Choose Y = N / √R. • Bounded subgraph Gi of size Ni has height √R, and thus can be partitioned as in case 1 to O(Ni / R) subgraphs, each of size O(R), using O((Ni /R)·√R) = O(Ni /√R) separator vertices. • In addition, the at most X cutter levels contribute another O(X · Y) = O(X · N/√R) separator vertices.

  20. Case 2: Analysis (2) • Thus, the total number of separator vertices is given by: S(N) ≤ O(X ·N/√R) + O(N/√R) + X · S(N/X) (and S(R) = 0) • Choose X = (M/B2)1/4. • For now, assume that R>B√M. • Thus X·N/√R = 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 > B2+ε.

  21. 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.

  22. 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/B+X·T(N/X), which is O(sort(N)). • Bounded height subgraphs can be separated “immediately”, using a total of O(sort(N)) I/Os.

  23. Case 2: I/O Efficiency (2) • So far we assumed that R > B√M. • If R ≤ B√M < 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/√R) separator vertices in each of the O(N/M) subgraphs, for a total of O(N/√R) vertices.

  24. Summery • We’ve 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.

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