Approximation and Hardness Results for Packing Cycles Mohammad R. Salavatipour Department of Computing Science University of Alberta Joint work with M. Krivelevich (Tel Aviv U.) Z. Nutov (Open U.) J. Verstraete (U. Waterloo) R. Yuster (U. Haifa)

2 Packing problems Example: Edge-disjoint path (EDP) problem Given a graph and a set of source-sink 10 T1T1 S2S2 S1S1 Pairs Goal: Find a maximum number of edge-disjoint s i,t i -paths. Classical and very well studied NP-hard problem. T2T2 S3S3 T3T3

3 EDP known results The problem has a large integrality gap even for planar graphs. 10 s1s1 t1t1 s2s2 t2t2 s3s3 t3t3 srsr trtr Fractional solution: Integral solution: 1

4 Directed graphs: Upper bounds: approx [GKRSY’03,KS04] Lower bounds: -hardness, unless [GKRSY’03] and (for DAGS) unless [MW’00] Undirected graphs: Upper bounds: has integrality gap [CKS’06] Lower bounds: -hardness unless [ACKZ’05] EDP known results:

5 Packing Disjoint Cycles What is the maximum number of edge-disjoint cycles (EDC) in a given graph ? What if the input graph is directed/undirected? Dual problems: For disjoint paths: multi-cut which has (for undirected) [GVY’96] and (for directed) [G’03] approx. For disjoint cycles: Feedback Arc/Vertex Set, -approx [S’95]

6 Results for EDC: EDC is APX-hard on undirected graphs and has approx [CPR’03] Theorem 1: For undirected graphs, a simple greedy gives -approx. Theorem 2: For directed graphs there is an -approx. Theorem 3: For directed graphs the problem is -hard, unless We can get -hardness unless

7 EDC on undirected graphs Theorem 1: For undirected graphs, a simple greedy gives -approx to the optimal fractional soluion. Algorithm: Repeat the following until G is empty: 1.Repeatedly, delete degree ≤ 1 vertices. 2.Repeatedly, short-cut every degree 2 vertex 3.Find and remove shortest cycle. 4.If go to step 1; else step3. v u w uw

8 Hardness of directed EDC Theorem 3: For directed graphs the EDC problem is -hard, unless First we show the following: Theorem 4: Directed EDC has an integrality gap of A natural attempt to prove integrality gap: use the grid construction for EDP

9 Integrality gap for directed EDC 10 Denote the grid graph with r pairs by D r Direct edges from top- to-bottom and left-to- right. Add a link from the sinks back to sources. The fractional solution still has size s1s1 t1t1 s2s2 t2t2 s3s3 t3t3 srsr trtr

10 Integrality gap for directed EDC (cont’d) 10 But there is also a large integral solution: We call these paths “non-canonical” or “cheating” paths. We have to make it costly to use cheating paths.

11 Integrality gap for directed EDC (cont’d) 10 Take two copies of the grid construction for EDP. Make graph H k as follows Observation: H k is acyclic. s1s1 t1t1 s2s2 t2t2 s3s3 t3t3 sksk tktk z1z1 z2z2 z3z3 zkzk HkHk We call the triple s i,t i,z i a block.

12 Integrality gap for directed EDC (cont’d) Consider H k and the 2k pairs s i,t i and s i,z i as an instance of EDP. Fact: Any optimal integral solution to EDP on H k either has one fully routed block or two partially routed block 10 s1s1 t1t1 s2s2 t2t2 s3s3 t3t3 sksk tktk z1z1 z2z2 z3z3 zkzk HkHk

13 Integrality gap for directed EDC (cont’d) 10 The idea is to start with several copies of the modified grid graph, say Also, take several copies of the graph s1s1 t1t1 s2s2 t2t2 s3s3 t3t3 srsr trtr We take a k -uniform r’- regular girth g hypergraph where

14 Integrality gap for directed EDC (cont’d) Every copy corresponds to a vertex of Every copy corresponds to an edge of Consider an arbitrary edge and let be its corresponding copy of Let’s call the blocks of, where block consist of triple with source and two sinks 10

15 Integrality gap for directed EDC (cont’d) 10 So we have k blocks Consider copies of corresponding to vertices of say s1s1 t1t1 srsr trtr Replace one intersection block in each with a block of

16 10 s1s1 t1t1 srsr trtr s1s1 t1t1 srsr Integrality gap for directed EDC (cont’d)

17 Integrality gap for directed EDC (cont’d) The k blocks of will be used to replace one intersection in the k copies, one from each Since is r’-regular, every belongs to r’ edges, so each of its intersection will be replaced with blocks of copies of from r’ different edges. Call this final graph

18 Integrality gap for directed EDC (cont’d) 10 s1s1 t1t1 srsr trtr Considering half integral solution and following canonical paths, every has r cycles with value ½, so fractional solution of has size s1s1 t1t1 srsr

19 Integrality gap for directed EDC (cont’d) 10 s1s1 t1t1 srsr trtr s1s1 t1t1 srsr The integral solution may have some canonical cycles and some cheating cycles (which will be long).

20 Integrality gap for directed EDC (cont’d) 1.Canonical cycles: Since each can allow at most two blocks be routed (partially or fully) and since each canonical cycle goes through r blocks there are at most short cycles with 10 s1s1 t1t1 srsr trtr 2.Long cycles: Each has edges; each long cycles uses g edges → long cycles Thus size of integral solution:

21 So the gap is: Choosing the parameters: We give explicit constructions for graph for which and Let r be constant and k=g; Then and gap = writing the in terms of n: Integrality gap for directed EDC (cont’d)

22 To turn this into a hardness proof we using the following result of [MW’00]: Theorem: Given a DAG G and source-sink pairs as an instance of EDP with it is quasi-NP-hard to decide: 1.All pairs can be routed 2.At most a fraction can be routed. The construction of will be based on two copies of the instance of EDP on DAGS. Hardness of directed EDC

23 The upper and lower bounds for directed EDC and undirected EDP are similar: Questions: What is the correct upper/lower bound for directed EDC? What about undir EDC? Conclusion Upper bound Lower bound Undir EDP Direct EDC Undir EDC