1. Approximating Maximum Subgraphs Without Short Cycles Guy Kortsarz Join work with Michael Langberg and Zeev Nutov

2. 2 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 subgraph of G of girth at least g with the maximum number of edges. g=4

3. 3 Max-g-Girth: context Max-g-Girth: Max-g-Girth: Used in study of “Genome Sequencing” [Pevzner Tang Tesler]. Used in study of “Genome Sequencing” [Pevzner Tang Tesler]. Mentioned in [Erdos Gallai Tuza] for g=4 (triangle free). Mentioned in [Erdos Gallai Tuza] for g=4 (triangle free). Complementary problem of “covering” all small cycles (size ≤ g) with minimum number of edges was studied in past. Complementary problem of “covering” all small cycles (size ≤ g) with minimum number of edges was studied in past. [Krivelevich] addressed g=4 (covering triangles). [Krivelevich] addressed g=4 (covering triangles). Approximation ratio of 2 was achieved (ratio of 3 is easy). Approximation ratio of 2 was achieved (ratio of 3 is easy). Problem is NP-Hard (even for g=4). Problem is NP-Hard (even for g=4).

4. Max-g-Girth on cliques The Max-g-Girth problem on cliques = densest graph on n vertices with girth g. The Max-g-Girth problem on cliques = densest graph on n vertices with girth g. Has been extensively studied [Erd¨os, Bondy Simonovits, …]]. Has been extensively studied [Erd¨os, Bondy Simonovits, …]]. Known that Max-g-Girth has size between  (n 1+4/(3g-12) ) and O(n 1+2/(g-2) ). Known that Max-g-Girth has size between  (n 1+4/(3g-12) ) and O(n 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 O(n -  ) will solve open problem. Implies that approximation ratio O(n -  ) will solve open problem. 4

5. 5 First steps Positive: Positive: Trivial by previous bounds approximation ratio of ~ n -2/(g-2). Trivial by previous bounds approximation ratio of ~ n -2/(g-2). For g=5,6  n -1/2. For g=5,6  n -1/2. If g>4 part of input: ratio 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|E G | edges  ratio ½. If g=4 (maximum triangle free graph): return random cut and obtain ½|E G | edges  ratio ½. g = 4: constant ratio, g ≥ 5 polynomial ratio! g = 4: constant ratio, g ≥ 5 polynomial ratio!

6. 6 Our results Max-g-Girth: positive and negative. Max-g-Girth: positive and negative. Positive: Positive: Improve on trivial n -1/2 for general g to n -1/3. Improve on trivial n -1/2 for general g to n -1/3. For g=4 (triangle free) improve from ½ to 2/3. For g=4 (triangle free) improve from ½ to 2/3. For instances with  n 2 edges: ratio ~  n -2/3g. For instances with  n 2 edges: ratio ~  n -2/3g. Negative: Negative: Max-g-Girth is APX hard (any g). Max-g-Girth is APX hard (any g). Large gap!

7. Our results Covering triangles by edges. Covering triangles by edges. [Krivelevich] presented LP based 2 approx. ratio. [Krivelevich] presented LP based 2 approx. ratio. Posed open problem of tightness of integrality gap. Posed open problem of tightness of integrality gap. We solve open problem: present family of graphs in which the gap is 2- . We solve open problem: present family of graphs in which the gap is 2- . Moreover: 2-  approximation implies 2-  for Vertex Cover (  <1/2). Moreover: 2-  approximation implies 2-  for Vertex Cover (  <1/2). 7

8. Positive Theorem: Max-g-Girth admits ratio ~ n -1/3. Theorem: Max-g-Girth admits ratio ~ n -1/3. Outline of proof: Outline of proof: Consider optimal subgraph H. Consider optimal subgraph H. Remove all odd cycles in G by randomly partitioning G and removing edges on each side. 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. ½ the edges of optimal H remain  Opt. value “did not” change. Now G is bipartite, need to remove even cycles of size < g. 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=5: only need to remove cycles of length 4. If g=6: only need to remove cycles of length 4. If g=6: only need to remove cycles of length 4. If g>6: as any graph of girth g=2r+1 or 2r+2 contains at most ~ n 1+1/r edges, trivial algorithm gives ratio n -1/3. If g>6: as any graph of girth g=2r+1 or 2r+2 contains at most ~ n 1+1/r edges, trivial algorithm gives ratio n -1/3. Goal: Approximate Max-5-Girth within ratio ~ n -1/3. Goal: Approximate Max-5-Girth within ratio ~ n -1/3. 8

9. Max-5-Girth Goal: App. Max-5-Girth on bipartite graphs within ratio ~ n -1/3. Goal: App. Max-5-Girth on bipartite graphs within ratio ~ n -1/3. Namely: given bipartite G find max. H  G without 4-cycles. Namely: given bipartite G find max. H  G without 4-cycles. Algorithm has 2 steps: Algorithm has 2 steps: Step I: Find G’  G that is almost regular (in both parts) such that Opt(G’)~Opt(G). 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 |E H | ≥ Opt(G’)n -1/3. Step II: Find H  G’ for which |E H | ≥ Opt(G’)n -1/3. 9 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).

10. Max-5-girth Goal: App. Max-5-girth on bipartite graphs within ratio ~ n -1/3. Goal: App. Max-5-girth on bipartite graphs within ratio ~ n -1/3. Namely: given bipartite G find max. H  G without 4-cycles. Namely: given bipartite G find max. H  G without 4-cycles. Algorithm has 2 steps: Algorithm has 2 steps: Step I: Find G’  G that is almost regular (in both parts) such that Opt(G’)~Opt(G). 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 |E H |≥ Opt(G’)n -1/3. Step II: Find H  G’ for which |E H |≥ Opt(G’)n -1/3. 10 Step II: Now G’ is regular. Enables us to tightly analyze the maximum amount of 4 cycles in G’. Regularity connects # of edges |E G’ | 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.

11. Covering k cycles Our algorithm actually gives an approximation for the problem of finding a maximum edge subgraph of G without cycles of length exactly k. Our algorithm actually gives an approximation for the problem of finding a maximum edge subgraph of G without cycles of length exactly k. Trivial algorithm (return spanning tree) gives ratio of n -2/k Trivial algorithm (return spanning tree) gives ratio of n -2/k Our algorithm gives ~ n -2/k (1+1/(k-1)) Our algorithm gives ~ n -2/k (1+1/(k-1)) Significant for small values of k. Significant for small values of k. 11

12. Some interesting open prob. LP for g=4 (maximum triangle free graph): LP for g=4 (maximum triangle free graph): Max:  e x(e) st: For every triangle C,  e  C x(e)  2 st: For every triangle C,  e  C x(e)  2 Max-Cut: integrality gap = 2. Max-Cut: integrality gap = 2. Complete graph: IG = 4/3 ( x(e)=2/3). Complete graph: IG = 4/3 ( x(e)=2/3). Conjecture: NP-Hard to obtain 2/3+  approx. Conjecture: NP-Hard to obtain 2/3+  approx. 12

13. Some interesting open prob. Max-5-girth: Large gap between upper and lower bounds. Large gap between upper and lower bounds. We suspect that for some  a ratio of n -  is NP-Hard. We suspect that for some  a ratio of n -  is NP-Hard. Obvious open problem: give strong lower bound for g=5. Obvious open problem: give strong lower bound for g=5. 13

14. Some interesting open prob. Set Cover in which each element appears in k sets. Upper bound: k. Upper bound: k. Lower bound: k-1-  [Dinur et al.] Lower bound: k-1-  [Dinur et al.] If sets are “k cycles” in given graph G we show a ratio of k-1 (for odd k). If sets are “k cycles” in given graph G we show a ratio of k-1 (for odd k). Open problem: is k-1 possible for general set cover (large k). Open problem: is k-1 possible for general set cover (large k). 14 Thanks!