Approximating Approximating Maximum Subgraphs Maximum Subgraphs

WithoutWithoutShort CyclesShort Cycles

Guy KortsarzGuy Kortsarz

Join work with Michael Langberg and Zeev Join work with Michael Langberg and Zeev

NutovNutov

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Max-g-GirthMax-g-Girth

Girth: A graph G is said to have girth g if its shortest cycle is of length g.

Max-g-Girth: Given G, find a subgraph of G of girth at least g with the maximum number of edges.

g=4

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Max-g-Girth: contextMax-g-Girth: context

•Max-g-Girth:Max-g-Girth:

•Used in study of “Genome Sequencing” Used in study of “Genome Sequencing” [Pevzner Tang [Pevzner Tang Tesler].Tesler].

•Mentioned in Mentioned in [Erdos Gallai Tuza][Erdos Gallai Tuza] for for g=4g=4 (triangle free). (triangle free).

• Complementary problemComplementary problem of “covering” all small cycles (size of “covering” all small cycles (size

≤ ≤ gg) with minimum number of edges was studied in past.) with minimum number of edges was studied in past.

•[Krivelevich][Krivelevich] addressed addressed g=4g=4 (covering triangles). (covering triangles).

•Approximation ratio of Approximation ratio of 22 was achieved (ratio of was achieved (ratio of 33 is is

easy).easy).

• Problem is NP-Hard (even for Problem is NP-Hard (even for g=4g=4).).

Max-g-Girth on cliquesMax-g-Girth on cliques

•The Max-g-Girth problem on cliques = densest graph on The Max-g-Girth problem on cliques = densest graph on

nn vertices with girth vertices with girth gg..

•Has been extensively studied Has been extensively studied [Erd¨os, Bondy Simonovits, …]][Erd¨os, Bondy Simonovits, …]]..

•Known that Max-g-Girth has size between Known that Max-g-Girth has size between (n(n1+4/(3g-12)1+4/(3g-12)))

and and O(nO(n1+2/(g-2)1+2/(g-2)))..

•There is a polynomial gap! Long standing open problem.There is a polynomial gap! Long standing open problem.

•Implies that approximation ratio Implies that approximation ratio O(nO(n--) ) will solve open will solve open

problem.problem.

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First stepsFirst steps

• PositivePositive: :

•Trivial by previous bounds approximation ratio of ~ Trivial by previous bounds approximation ratio of ~ nn-2/(g-2)-2/(g-2)..

•For For g=5,6 g=5,6 nn-1/2-1/2..

•If If g>4g>4 part of input: ratio part of input: ratio nn-1/2-1/2..

•If If g=4g=4 (maximum triangle free graph): return random cut (maximum triangle free graph): return random cut

and obtain and obtain ½|E½|EGG| | edgesedges ratioratio ½. ½.

•g = 4g = 4: constant ratio, : constant ratio, g ≥ 5g ≥ 5 polynomial ratio! polynomial ratio!

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Our resultsOur results

•Max-g-Girth: Max-g-Girth: positive and negative.positive and negative.

• PositivePositive: :

•Improve on trivial Improve on trivial nn-1/2-1/2 for general for general g g toto n n-1/3-1/3. .

•For For g=4g=4 (triangle free) improve from (triangle free) improve from ½½ to to 2/3.2/3.

•For instances withFor instances with nn22 edges: ratio ~ edges: ratio ~ nn-2/3g-2/3g..

•NegativeNegative::

•Max-g-Girth is APX hard (any Max-g-Girth is APX hard (any gg).).

Large gap!

Our resultsOur results

•Covering triangles by edges.Covering triangles by edges.

•[Krivelevich] presented LP based [Krivelevich] presented LP based 22 approx. ratio. approx. ratio.

•Posed open problem of tightness of integrality Posed open problem of tightness of integrality

gap.gap.

•We solve open problem: present family of graphs We solve open problem: present family of graphs

in which the gap is in which the gap is 2-2-..

•Moreover: Moreover: 2-2- approximation implies approximation implies 2-2- for for

Vertex Cover Vertex Cover ((<1/2)<1/2)..

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PositivePositive• Theorem: Theorem: Max-g-Girth admits ratio ~Max-g-Girth admits ratio ~ n n-1/3-1/3..

•Outline of proof:Outline of proof:

•Consider optimal subgraphConsider optimal subgraph HH..

•Remove all odd cycles in Remove all odd cycles in GG by randomly partitioning by randomly partitioning GG and and removing edges on each side.removing edges on each side.

•½½ the edges of optimal the edges of optimal HH remain remain Opt. value “did not” Opt. value “did not” change.change.

•Now Now GG is bipartite, need to remove is bipartite, need to remove eveneven cycles of size < cycles of size < gg..

•If If g=5g=5: only need to remove cycles of length : only need to remove cycles of length 44..

•If If g=6g=6: only need to remove cycles of length : only need to remove cycles of length 44..

•If If g>6g>6: as: as any graph of girth any graph of girth g=2r+1 g=2r+1 oror 2r+2 2r+2 contains at most contains at most ~ ~ nn1+1/r1+1/r edges, trivial algorithm gives ratio edges, trivial algorithm gives ratio nn-1/3-1/3..

•GoalGoal: Approximate Max-5-Girth within ratio ~: Approximate Max-5-Girth within ratio ~ n n-1/3-1/3..8

Max-5-GirthMax-5-Girth

•GoalGoal: App. Max-5-Girth on bipartite graphs within ratio ~: App. Max-5-Girth on bipartite graphs within ratio ~ n n--

1/31/3..

•NamelyNamely: given bipartite : given bipartite GG find max. find max. HHGG without without 44-cycles.-cycles.

• Algorithm has 2 steps:Algorithm has 2 steps:

•Step IStep I: Find : Find G’G’G G that is almost regular (in both parts) that is almost regular (in both parts) such that such that Opt(G’)~Opt(G)Opt(G’)~Opt(G)..

•Step IIStep II: Find : Find HHG’ G’ for which for which |E|EHH| ≥ Opt(G’)n| ≥ Opt(G’)n-1/3-1/3..

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Step I: •General procedure that may be useful elsewhere.•Let G=(A,B;E) – want G’ almost regular on A and B & Opt(G’)~Opt(G).•Starting point: easy to make A regular (bucketing).•Now we can make B regular, however A becomes irregular.•Iterate … •Can show: if we do not converge after constant # steps then it must be the case that Opt(G) is small (in each iteration degree decreases).

Max-5-girthMax-5-girth

•GoalGoal: App. Max-5-girth on bipartite graphs within ratio ~: App. Max-5-girth on bipartite graphs within ratio ~ n n--

1/31/3..

•NamelyNamely: given bipartite : given bipartite GG find max. find max. HHGG without without 44-cycles.-cycles.

• Algorithm has 2 steps:Algorithm has 2 steps:

•Step IStep I: Find : Find G’G’G G that is almost regular (in both parts) that is almost regular (in both parts) such that such that Opt(G’)~Opt(G)Opt(G’)~Opt(G)..

•Step IIStep II: Find : Find HHG’ G’ for which for which |E|EHH|≥ Opt(G’)n|≥ Opt(G’)n-1/3-1/3..

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Step II: •Now G’ is regular.•Enables us to tightly analyze the maximum amount of 4 cycles in G’.

Regularity connects # of edges |EG’| with number of 4-cycles.•Remove edges randomly as to break 4-cyles (“alteration method”).•Using comb. upper bound on Opt [NaorVerstraete] yields n-1/3 ratio.

Covering k cyclesCovering k cycles

•Our algorithm actually gives an approximation for Our algorithm actually gives an approximation for

the problem of finding a maximum edge the problem of finding a maximum edge

subgraph of subgraph of GG without cycles of length exactly without cycles of length exactly kk..

•Trivial algorithm (return spanning tree) gives Trivial algorithm (return spanning tree) gives

ratio of ratio of nn-2/k-2/k

•Our algorithm gives ~ Our algorithm gives ~ nn-2/k (1+1/(k-1)) -2/k (1+1/(k-1))

• Significant for small values of Significant for small values of kk..

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Some interesting open Some interesting open prob.prob.

•LP for LP for g=4g=4 (maximum triangle free graph): (maximum triangle free graph):

Max:Max: eex(e)x(e)

st:st: For every triangle C, For every triangle C, eeCCx(e)x(e)22

•Max-Cut: integrality gap = Max-Cut: integrality gap = 22..

•Complete graph: IG = Complete graph: IG = 4/3 (4/3 (x(e)=2/3)x(e)=2/3)..

•Conjecture: NP-Hard to obtain Conjecture: NP-Hard to obtain 2/3+2/3+ approx. approx.

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Some interesting open Some interesting open prob.prob.Max-5-girth:Max-5-girth:

•Large gap between upper and lower Large gap between upper and lower

bounds.bounds.

•We suspect that for some We suspect that for some a ratio of a ratio of nn-- isis

NP-Hard.NP-Hard.

•Obvious open problem: give strong lower Obvious open problem: give strong lower

bound for bound for g=5g=5..13

Some interesting open Some interesting open prob.prob.Set Cover in which each element appears in Set Cover in which each element appears in kk

sets.sets.

•Upper bound: Upper bound: kk..

•Lower bound: Lower bound: k-1-k-1- [Dinur et al.][Dinur et al.]

•If sets are “k cycles” in given graph G we show a If sets are “k cycles” in given graph G we show a

ratio of ratio of k-1k-1 (for odd (for odd kk).).

•Open problem: is Open problem: is k-1k-1 possible for general set cover possible for general set cover

(large (large kk).).

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Thanks!