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Two subgraph maximization problems - PowerPoint PPT Presentation

Two subgraph maximization problems. Michael Langberg. Open University Israel. Joint with Guy Kortsarz, Zeev Nutov, Yuval Rabani and Chaitanya Swamy. This talk: overview. Two maximization problems. Not addressed in the past (in this context). Part I : Subgraph homomorphism .

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Two subgraph maximization problems

Michael Langberg

Open University Israel

Joint with Guy Kortsarz, Zeev Nutov, Yuval Rabani and ChaitanyaSwamy

• Two maximization problems.

• Not addressed in the past (in this context).

• Part I: Subgraph homomorphism.

• Part II: Subgraphs with large girth.

Part I:Subgraph Homomorphism

Homomorphism:

(u,v)EG: ((u),(v))EH

If H is a k-clique:

H-coloring  k-coloring

• A k-coloring of G is an assignment of colors the vertices of G such that each edge is adjacent to different colors.

• H-coloring (extends k-coloring):

• Input: Graphs G=(VG,EG) and H=(VH,GH).

• Output: Mapping : VGVH.

• Objective: All edges of G aremapped to edges of H.

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G

H

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• H-coloring is a decision problem.

• Study a maximization version of H-coloring.

• Maximum Graph Homomorphism (MGH).

• Present both upper and lower bounds.

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2

G

H

Max (u,v)EG s.t. ((u),(v))EH

• MGH:

• Input: Graphs G=(VG,EG) and H=(VH,GH).

• Output: Mapping : VGVH.

• Objective:Max. # edges of G mapped to edges of H.

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G

H

• Generalizes classical optimization problems:

• Max-Cut: H is a single edge.

G

H

• Generalizes classical optimization problems:

• Max-Cut: H is a single edge.

• Max-k-Cut: H is a k clique.

• MGH has not been addressed directly in the past.

• Related problems:

• H coloring (decision, counting) [HellNesteril,DyerGreenhill, BorgsChayesLovaszSosVesztergombi,CooperDyerFrieze, DyerGoldbergJerrum …].

• Maximum common subgraph [Kann].

• Minimum graph homomorphism [CohenCooperJeavonsKrokhin, AggarwalFederMotwaniZhu,GutinRafieyYeoTso].

• Positive:

• Easy to obtain ½ approximation.

Reduce to Max-Cut (map all of G to one edge in H).

• Easy to obtain (k-1)/k if H contains k clique.

• Negative:

• Cannot do better than Max-Cut (no PTAS):

• 0.878 unless UGC is false [KhotKindlerMosselO’Donnell].

Based on algorithm for “light Max-Cut” analyzed in [CharikarWirth] using SDP and RPR2 rounding [FeigeL]

• MGH: both positive and negative.

• Positive:

• Improve on ½ when H is of constant size:

Ratio = ½ + (1/|VH|log|VH|)

• Negative:

• For general H, cannot improve on ½ unless random subgraph isomorphism P.

• Theorem: Cannot app. MGH within ratio > ½ unless“random subgraph isomorphism” P.

• Consider general G and H:

• “Subgraph Isomorphism”: IsGa subgraph ofH?

• NP-hard (e.g., encodes Hamiltonian cycle).

• What happens if G and H are chosen from a certain distribution over graphs?

G

H

• Consider graphs G and H in which

• Vertex sets are of size n.

• Chosen from Gn,p.

• We take ½ > pH>> pG > log(n)/n.

• Is Sub. Isomorphism solvable on such instances (w.h.p.)?

• Not hard to verify:

• W.h.p. a random G will not be a subgraph of a random H.

• So “random subgraph isomorphism” is solvable in P (w.h.p.).

• What can we use as a hardness assumption?

G

H

• Theorem: Approximating MGH beyond ½ is as hard as the following refutation problem.

• Design an algorithm that given random G and H:

• If GH algorithm must return “yes” answer.

• Algorithm must return “no” answer with prob. > ½.

• Refutation algorithms have been studied in the context of approximation algorithms [Feige,Alekhnovich,Demaine et al.,…].

G

H

G

H

• Theorem: Approximating MGH beyond ½ is as hard as the following refutation problem.

• Design an algorithm that given random G and H:

• If GH algorithm must return “yes” answer.

• Algorithm must return “no” answer with prob. > ½.

• Main Lemma: W.h.p. over random G and H:

• MGH(G,H) ≤ |EG|(½+o(1))

• Suffices to prove Theorem:

• Assume MGH has algorithm with ratio ½+.

• If GH: MGH(G,H)=|EG|  approx. will give |EG|(½+)  “yes”.

• By Lemma:with prob. > ½: MGH(G,H) ≤ |EG|(½+o(1))  “no”.

• Proof of lemma: Need a different distribution then previously def.

• Need H (and thus G) to be random and triangle free.

• G and H:

• Chosen from Gn,p.

• We take pH>> pG > log(n)/n.

• Removing edges for -free.

• Theorem: Approximating MGH beyond ½ is as hard as the following refutation problem.

• Good question!

• Techniques used for Graph Isomorphism seem to fail.

• Local analysis seems to fail.

• May make assumption more robust (require “yes” even if mapping captures many edges of G).

• Summary I:

• Ratio ½ + 1/|VH|log|VH|.

• “Hard” to improve ½.

?

• G and H:

• Chosen from Gn,p.

• We take pH>> pG > log(n)/n.

• Removing edges for -free.

G

H

Part II Graphs with large 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 maximum subgraph of G with girth at least g.

g=4

• Max-g-Girth has not been addressed in the past.

• Mentioned in [ErdosGallaiTuza] for g=4 (triangle free).

• Used in study of “Genome Sequencing” [PevznerTangTesler].

• Complementary problem of “covering” all small cycles (size ≤ g) with minimum number of edges was studied in past.

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

• Approximation ratio of 2 was achieved (ratio of 3 is easy).

• Problem is NP-Hard (even for g=4).

• Positive:

• Any graph of girth g=2r+1 or 2r+2 contains at most ~ n1+1/r edges (girth g  O(n1+2/g) edges) [AlonHooryLinial].

• A spanning tree of G results in app. ratio of ~ n-1/r = n-2/g.

• Polynomial approximation: g=4 n-1;g=5,6  n-1/2.

• If g>4 part of input: ratio n-1/2.

• If g=4 (maximum triangle free graph): return random cut and obtain ½|EG| edgesratio ½.

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

• Max-g-Girth: positive and negative.

• Positive:

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

• For g=4 (triangle free) improve from ½ to 2/3.

• Negative:

• Max-g-Girth is APX hard (any g).

• Proof of positive result for g=4 uses ratio of 2 obtained by [Krivelevich] on complementary problem of “covering” triangles.

• We show that result of [Krivelevich] is bestpossible unless Vertex Cover can be approximated within ratio < 2.

Large gap!

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

• Outline of proof:

• Consider optimal subgraphH.

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

• ½ the edges of optimal H remain  Opt. value “did not” change.

• Now G is bipartite, need to remove even cycles of size < g.

• If g=5: only need to remove cycles of length 4.

• If g=6: only need to remove cycles of length 4.

• If g>6: asany graph of girth g=2r+1 or 2r+2 contains at most ~ n1+1/r edges, trivial algorithm gives ratio n-1/3.

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

Step I:

• General procedure that may be useful elsewhere.

• “Iterative bucketing”.

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.

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• Goal: App. Max-5-Girth on bipartite graphs within ratio ~ n-1/3.

• Namely: given bipartite G find max. HG without 4-cycles.

• Algorithm has 2 steps:

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

• Step II: Find HG’ for which |EH| ≥ Opt(G’)n-1/3.

• Studied two max. subgraph problems:

• Maximum Graph Homomorphism

• Max-g-Girth.

• Open questions:

• MGH:

• Base hardness of app. on standard assumptions.

• Refuting Subgraph Isomorphism vs. refuting Max-Sat.

• Max-g-Girth:

• Polynomial gap between upper and lower bounds (g=5 especially appealing).

Thanks!