Shortest Non-trivial Cycles in Directed and Undirected Surface - - PowerPoint PPT Presentation

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Shortest Non-trivial Cycles in Directed and Undirected Surface - - PowerPoint PPT Presentation

Shortest Non-trivial Cycles in Directed and Undirected Surface Graphs Kyle Fox University of Illinois at Urbana-Champaign Surfaces 2-manifolds (with boundary) genus g : max # of disjoint simple cycles whose compliment is connected =


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Shortest Non-trivial Cycles in Directed and Undirected Surface Graphs

Kyle Fox

University of Illinois at Urbana-Champaign

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Surfaces

  • 2-manifolds (with boundary)
  • genus g: max # of disjoint simple cycles

whose compliment is connected = number of holes = number of handles attached to sphere

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Surface Graphs

  • n vertices as points
  • m edges as (mostly) disjoint curves
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Surface Graphs

  • n vertices as points
  • m edges as (mostly) disjoint curves
  • Assume g = O(n) and m = O(n)
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Surface Graphs

  • n vertices as points
  • m edges as (mostly) disjoint curves
  • Assume g = O(n) and m = O(n)
  • We want to find non-trivial cycles
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Non-trivial Cycles

Non-separating Non-contractible Trivial ಠ_ಠ

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Finding Short Non-trivial Cycles

  • Want to minimize sum of real edge lengths

(not geodesics)

  • Natural question for surface embedded

graphs

  • Cutting along non-trivial cycles reduces the

complexity of the graph

  • Useful for combinatorial optimization,

graphics, graph drawing, …

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90]

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90] O(n2 log n) O(n2 log n) [Erickson, Har-peled ’04]

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90] O(n2 log n) O(n2 log n) [Erickson, Har-peled ’04] gO(g) n3/2 O(g3/2 n3/2 log n + g5/2 n1/2) [Cabello, Mohar ’07]

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90] O(n2 log n) O(n2 log n) [Erickson, Har-peled ’04] gO(g) n3/2 O(g3/2 n3/2 log n + g5/2 n1/2) [Cabello, Mohar ’07] gO(g) n log n gO(g) n log n [Kutz ’06]

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90] O(n2 log n) O(n2 log n) [Erickson, Har-peled ’04] gO(g) n3/2 O(g3/2 n3/2 log n + g5/2 n1/2) [Cabello, Mohar ’07] gO(g) n log n gO(g) n log n [Kutz ’06] O(g2 n log n) O(g2 n log n) [Cabello, Chambers ’06; C, C, Erickson ’12]

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90] O(n2 log n) O(n2 log n) [Erickson, Har-peled ’04] gO(g) n3/2 O(g3/2 n3/2 log n + g5/2 n1/2) [Cabello, Mohar ’07] gO(g) n log n gO(g) n log n [Kutz ’06] O(g2 n log n) O(g2 n log n) [Cabello, Chambers ’06; C, C, Erickson ’12] gO(g) n log log n gO(g) n log log n [Italiano, et al. ’11]

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90] O(n2 log n) O(n2 log n) [Erickson, Har-peled ’04] gO(g) n3/2 O(g3/2 n3/2 log n + g5/2 n1/2) [Cabello, Mohar ’07] gO(g) n log n gO(g) n log n [Kutz ’06] O(g2 n log n) O(g2 n log n) [Cabello, Chambers ’06; C, C, Erickson ’12] gO(g) n log log n gO(g) n log log n [Italiano, et al. ’11] 2O(g) n log log n 2O(g) n log log n [F ’13]

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Results (Undirected)

Non-con. Non-sep. O(n3) O(n3) [Thomassen ’90] O(n2 log n) O(n2 log n) [Erickson, Har-peled ’04] gO(g) n3/2 O(g3/2 n3/2 log n + g5/2 n1/2) [Cabello, Mohar ’07] gO(g) n log n gO(g) n log n [Kutz ’06] O(g2 n log n) O(g2 n log n) [Cabello, Chambers ’06; C, C, Erickson ’12] gO(g) n log log n gO(g) n log log n [Italiano, et al. ’11] 2O(g) n log log n 2O(g) n log log n [F ’13]

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New Undirected Results

  • Based on algorithm of Kutz [’06] and

Italiano et al. [’11]

  • Two cycles are homotopic if one can be

continuously deformed to the other

  • Prior algorithms compute shortest cycles in

gO(g) homotopy classes

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New Undirected Results

  • New algorithm reduces number of

homotopy classes to 2O(g)

  • Classes chosen based on triangulations of

the polygonal schema [Chambers et al. ’08; Chambers, Erickson, Nayyeri ’09]

1 1 1 1

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Undirected Edges are Kind

  • Walks have the same length as their

reversals

  • Shortest paths cross at most once
  • Neither holds in general for directed

graphs

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Results (Directed)

Non-con. Non-sep. O(n2 log n) and O(g1/2 n3/2 log n) O(n2 log n) and O(g1/2 n3/2 log n) [Cabello, Colin de Verdière, Lazarus ’10]

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Results (Directed)

Non-con. Non-sep. O(n2 log n) and O(g1/2 n3/2 log n) O(n2 log n) and O(g1/2 n3/2 log n) [Cabello, Colin de Verdière, Lazarus ’10] 2O(g) n log n [Erickson, Nayyeri ’11]

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Results (Directed)

Non-con. Non-sep. O(n2 log n) and O(g1/2 n3/2 log n) O(n2 log n) and O(g1/2 n3/2 log n) [Cabello, Colin de Verdière, Lazarus ’10] 2O(g) n log n [Erickson, Nayyeri ’11] gO(g) n log n O(g2 n log n) [Erickson ’11]

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Results (Directed)

Non-con. Non-sep. O(n2 log n) and O(g1/2 n3/2 log n) O(n2 log n) and O(g1/2 n3/2 log n) [Cabello, Colin de Verdière, Lazarus ’10] 2O(g) n log n [Erickson, Nayyeri ’11] gO(g) n log n O(g2 n log n) [Erickson ’11] O(g3 n log n) [F ’13]

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Results (Directed)

Non-con. Non-sep. O(n2 log n) and O(g1/2 n3/2 log n) O(n2 log n) and O(g1/2 n3/2 log n) [Cabello, Colin de Verdière, Lazarus ’10] 2O(g) n log n [Erickson, Nayyeri ’11] gO(g) n log n O(g2 n log n) [Erickson ’11] O(g3 n log n) [F ’12]

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Assumptions for New Result

  • If the shortest non-contractible cycle is

separating, we can use the algorithm of Erickson

  • Presentation assumes the cycle is

separating and the surface has exactly one boundary cycle

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Main Ideas

  • Lift the graph to one of O(g) copies of a

covering space

  • The shortest non-contractible cycle is non-

null-homologous in one of the lifted copies

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Non-null-homologous Cycles

  • Either non-separating or separate boundary

components

  • Are all non-contractible
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Non-null-homologous Cycles

  • Bonus Result: Shortest non-null-

homologous cycles computable as quickly as shortest non-separating cycles

  • 2O(g) n log log n time in undirected graphs
  • O(g2 n log n) time in directed graphs
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Covering Spaces

… …

  • Each point x in the original space lies in an
  • pen neighborhood U such that one or

more open neighborhoods in the covering space have a homeomorphism to U

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Covering Spaces

… …

  • Each walk in the covering space projects to

a walk in the original space

  • Any walk on the original surface has at

most one lift to the covering space that begins on a particular point

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Infinite Cyclic Cover

  • Let λ be any non-separating cycle
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Infinite Cyclic Cover

  • Cut the surface along λ
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Infinite Cyclic Cover

… …

  • Cut the surface along λ
  • Glue an infinite number of copies together

along λ

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Cycles in the Cover

… …

  • A cycle γ on the original surface lifts to a

path

  • Endpoints of path describe number of times

γ passes λ left-to-right and right-to-left

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Cycles in the Cover

… …

  • Cycle γ on the original surface lifts to a

cycle if and only if it crosses λ left-to-right the same number of times as it crosses right-to-left

  • Any separating cycle lifts to a cycle
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Cycles in the Cover

… …

  • The shortest non-contractible cycle lifts to

a cycle of the same length = =

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Path Intersections

… …

  • The shortest non-contractible cycle

intersects at most 2 lifts of any shortest path [Erickson ’11] = =

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Restricted Infinite Cyclic Cover

  • We only need 5 copies of the original

surface in the cover if we cut along a cycle made from two shortest paths

  • Leave boundaries at the ends of the left and

rightmost copies = =

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Non-contractible Lift

  • A cycle γ’ in the cover projects to a non-

contractible cycle if and only if γ’ is non- contractible No!

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Non-contractible Lift

  • The shortest non-contractible cycle in the
  • riginal surface is the shortest non-

contractible cycle in the cover = =

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Recap

  • Many non-separating cycles can be used to

create the restricted infinite cyclic cover

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Recap

  • Many non-separating cycles can be used to

create the restricted infinite cyclic cover

  • Suffices to find the shortest non-

contractible cycle in any subset of the cover

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Recap

  • Many non-separating cycles can be used to

create the restricted infinite cyclic cover

  • Suffices to find the shortest non-

contractible cycle in any subset of the cover

  • But the genus increased!
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Separating Boundary

  • Compute a system of 2g

non-separating cycles from shortest paths in O(n log n + g n) time (a homology basis) =

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Separating Boundary

  • Compute a system of 2g

non-separating cycles from shortest paths in O(n log n + g n) time (a homology basis) = =

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Separating Boundary

  • Compute a system of 2g

non-separating cycles from shortest paths in O(n log n + g n) time (a homology basis) = =

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Separating Boundary

  • Compute a system of 2g

non-separating cycles from shortest paths in O(n log n + g n) time (a homology basis) = =

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Separating Boundary

  • At least one of the non-separating cycles

yields a useful copy of the restricted infinite cyclic cover = =

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Separating Boundary

  • Main Lemma: At least one lift of the

shortest non-contractible cycle is a shortest non-null-homologous cycle

  • Lift is non-separating or separates a pair of

boundary = =

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Separating Boundary

  • Intuition: the shortest non-contractible

cycle separates the boundary component from a surface subset containing genus

  • Genus becomes boundary in the restricted

cover = =

Boundary Genus

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Separating Boundary

  • Theory: A non-separating cycle ω is

separated from the boundary

  • ω lifts to an arc separated from lift of
  • riginal boundary

= =

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Algorithm: Search the Covers

  • In each copy of the restricted infinite cyclic

cover, find the shortest cycle that is non- null-homologous = = ?

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Running Time

  • Can search for short cycles in

O(g2 n log n) time per covering space

  • O(g3 n log n) time spent searching 2g

covers

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Summary of Results

  • New algorithms for computing shortest

non-trivial cycles in undirected and directed surface graphs

  • Included O(g3 n log n) time algorithm for

shortest non-contractible cycles in directed graphs – first with near-linear dependency

  • n n and sub-exponential dependency on g
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Open Problems

  • Can we do O(g2 n log n)?
  • Would be easy if there always existed

some lift separating boundary

  • Other problems in directed graphs
  • Minimum s,t-cut
  • Minimum quotient cut
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Thank you