Two subgraph maximization problems

1. Two subgraph maximization problems Michael Langberg Open University Israel Joint with Guy Kortsarz, Zeev Nutov, Yuval Rabani and ChaitanyaSwamy

2. This talk: overview • Two maximization problems. • Not addressed in the past (in this context). • Part I: Subgraph homomorphism. • Part II: Subgraphs with large girth.

3. Part I:Subgraph Homomorphism

4. 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

5. 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.

6. 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.

7. Maximum Graph Homomorphism 4 1  3 2 G H • Generalizes classical optimization problems: • Max-Cut: H is a single edge.

8. Maximum Graph Homomorphism  G H • Generalizes classical optimization problems: • Max-Cut: H is a single edge. • Max-k-Cut: H is a k clique.

9. 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].

10. 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].

11. 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.

12. 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

13. 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

14. 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

15. 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.

16. 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

17. Part II Graphs with large girth

18. 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

19. 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).

20. 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!

21. 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!

22. 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.

23. 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.

24. 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!