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Two subgraph maximization problems

Michael Langberg

Open University Israel

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

This talk: overview

- Two maximization problems.
- Not addressed in the past (in this context).
- Part I: Subgraph homomorphism.
- Part II: Subgraphs with large girth.

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 : VGVH.
- Objective: All edges of G aremapped to edges of H.

4

1

G

H

3

2

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

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 : VGVH.
- Objective:Max. # edges of G mapped to edges of H.

Maximum Graph Homomorphism

4

1

3

2

G

H

- Generalizes classical optimization problems:
- Max-Cut: H is a single edge.

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 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

- 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

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

- 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

- 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

- Theorem: Approximating MGH beyond ½ is as hard as the following refutation problem.
- Design an algorithm that given random G and H:
- If GH 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 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 GH 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 GH: 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?

- 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

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 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 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| edgesratio ½.
- g = 4: constant ratio, g ≥ 5 polynomial ratio!

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

- 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

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. HG 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 HG’ for which |EH| ≥ Opt(G’)n-1/3.

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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