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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
Two subgraph maximization problems

Michael Langberg

Open University Israel

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


This talk overview
This talk: overview

  • 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

Part I:Subgraph Homomorphism


K coloring and h coloring
k-coloring and H-coloring

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.

4

1

G

H

3

2


Part i
Part I

  • H-coloring is a decision problem.

  • Study a maximization version of H-coloring.

  • Maximum Graph Homomorphism (MGH).

  • Present both upper and lower bounds.


Maximum graph homomorphism
Maximum Graph Homomorphism

4

1

3

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.


Maximum graph homomorphism1
Maximum Graph Homomorphism

4

1

3

2

G

H

  • Generalizes classical optimization problems:

    • Max-Cut: H is a single edge.


Maximum graph homomorphism2
Maximum Graph Homomorphism

G

H

  • Generalizes classical optimization problems:

    • Max-Cut: H is a single edge.

    • Max-k-Cut: H is a k clique.


Mgh context
MGH: context

  • 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].


First steps
First steps

  • 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):

      • 16/17 unless P=NP [Hastad].

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


Our results 1
Our results #1

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.


Negative result
Negative Result

  • 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


Random instances
Random instances

  • 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


Hardness assumption
Hardness assumption

  • 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


Hardness assumption1
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.

  • 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.


Is the assumption strong
Is the assumption strong?

  • 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

Part II Graphs with large girth


Max g girth
Max-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 maximum subgraph of G with girth at least g.

g=4


Max g girth context
Max-g-Girth: context

  • 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).


First steps1
First steps

  • 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!


Our results 2
Our results #2

  • 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!


Positive
Positive

  • 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.


Max 5 girth
Max-5-Girth

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.

25

  • 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.


Concluding remarks
Concluding remarks

  • 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!


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