1. Iterative Rounding in Graph Connectivity Problems Kamal Jain ex- Georgia Techie Microsoft Research Some slides borrowed from Lap Chi Lau.

2. The Main Object A graph G = (V,E); A connectivity requirement r(u,v) for each pair of vertices u,v. Steiner Network A subgraph H of G which has r(u,v) disjoint paths between each pair u,v. undirected or directed edge or vertex

3. Examples of Steiner Network Spanning tree : r(u,v) = 1 for all pairs of vertices. Steiner tree : r(u,v) = 1 for all pair of required vertices. Steiner forest : r(s i,t i ) = 1 for all source sink pairs. k-edge-connected subgraph : r(u,v) = k for all pairs of vertices.

4. Survivable Network Design Survivable network design : find a “good” Steiner network Minimum cost Steiner network: Given a cost c(e) on each edge, find a Steiner network with minimum total cost. Minimum degree Steiner network: Find a Steiner network with minimum maximum degree. e.g. minimum spanning tree, minimum Steiner tree e.g. Hamiltonian path, Hamiltonian cycle

5. Three Kinds of Problems Only the first objective: Minimize the cost of the edges. Only the second objective: Minimize the maximum degree. Both objectives: Minimize the cost of the edges as well as the maximum degree.

6. Minimum cost Steiner Network Given A an undirected graph G = (V,E), Cost c(e) on edges, and A connectivity requirement r(u,v) for each pair of vertices u,v. Feasibility A subgraph H of G which has r(u,v) edge disjoint paths between each pair u,v. Objective Minimize the cost of H. Result: Factor 2 approximation algorithm.

7. Linear Programming Relaxation S uv Write f(S) := max{ r(u,v) | S separates u and v}. At least r(u,v) edges crossing S

8. Linear Programming Solver Suitable Rounding Procedure Problem Instance Optimal Fractional Solution Integer Solution Linear Programming Solver Problem Instance Optimal Fractional Solution Part Integer Good Part Too much Fractional Residual Problem Typical Rounding: Iterative Rounding:

9. Linear Programming Relaxation Theorem [J]. Every basic solution has an edge with value at least 1/2 A half edge

10. Iterative Rounding Initialization: H =, f’ = f. While f’ ≠ 0 do: oFind a basic optimal solution, x, of the LP with function f’. oAdd an edge with x(e) ≥ 1/2 into H. oUpdate f’: for every set S, set Output H. By previous Theorem 0.5 e f’(S)=2 f’(S)=1 Corollary. This is a 2-approximation algorithm for the minimum cost Steiner network problem. The residual problem is feasible.

11. A Proof Sketch An edge of 0, delete it. An edge of 1, pick it. Tight inequalities all come from the connectivity requirements. A basic solution is the unique solution of m linearly independent “tight” constraints, where m is the number of variables in the LP.

12. A Proof Sketch Uncrossing technique : uncross “intersecting” sets A B A[BA[B AÅBAÅB Each tight constraint corresponds to a set of vertices in the graph.

13. A Proof Sketch If there are no half-edges, derive a contradiction by a counting argument. no intersecting tight sets => number of linearly independent tight constraints is small no half-edges => number of edges is large In a basic solution, number of linearly independent tight constraints = number of edges Laminar family 1/3 Contradiction!

14. Iterative Rounding Theorem [J]. Iterative rounding gives factor 2 approximation algorithm. a b Cost ≤ 2a ≤b Re-Optimize ≤2a ≤2b Integer Theorem Induction

15. One shot rounding is hard e.g., Petersen Graph Use every edge to the extent of 1/3.

16. Basic Feasible Solution

17. Tight sets

18. The Problem Statement with both objectives Goal: to find a good Steiner network w.r.t. to both criterion Minimum cost Steiner network with degree constraints: Given a cost c(e) on each edge, find a Steiner network with minimum total cost, so that every vertex has degree at most B. Without degree bounds, this is the minimum cost Steiner network problem. Without cost on edges, this is the minimum degree Steiner network problem.

19. Ideal Approximation Minimum cost Steiner network with degree constraints: Given a cost c(e) on each edge, find a Steiner network with minimum total cost, so that every vertex has degree at most B. Let OPT(B) be the value of an optimal solution to this problem. Ideally, we would like to return a solution so that: SOL(B) ≤ c·OPT(B) However, it cannot be done for any polynomial factor, even for B=2, since this generalizes the minimum cost Hamiltonian path problem.

20. Bicriteria Approximation Algorithms This implies a c-approximation for minimum cost Steiner network, and a f(B)-approximation for minimum degree Steiner network. Minimum cost Steiner network with degree constraints: Given a cost c(e) on each edge, find a Steiner network with minimum total cost, so that every vertex has degree at most B. Let OPT(B) be the value of an optimal solution to this problem. A (c,f(B))-approximation algorithm if it returns a solution with SOL(f(B)) ≤ c·OPT(B) maximum degree f(B)e.g. f(B)=2B+10

21. Results without and with Iterative Rounding Minimum costMinimum degreeBicriteria Spanning tree1B+1 [FR](1, B+2) [G] Steiner tree1.55 [RZ]B+1 [FR](O(logn),O(logn)B) Steiner forest2 [AKR]?? k-ec subgraph2 [KV]O(log n)·B [FMZ]? Steiner network2 [J]?? Main Theorem [LNSS]: There is a (2,2B+3)-approximation algorithm for the minimum Steiner network problem with degree constraints. Corollary : There is a constant factor approximation algorithm for the Minimum Degree Steiner Network problem. (2,2B+3) 2B+3 (1, B+1)

22. Linear Programming Relaxation S uv Write f(S) := max{ r(u,v) | S separates u and v}. At least r(u,v) edges crossing S Nonuniform degree bounds

23. Linear Programming Relaxation Nonuniform degree bounds Theorem [J]. Every basic solution has an edge with value at least 1/2 A half edge

24. Iterative Rounding Initialization: H =, f’ = f. While f’ ≠ 0 do: oFind a basic optimal solution, x, of the LP with function f’. oAdd an edge with x(e) ≥ 1/2 into H. oUpdate f’: for every set S, set Output H. By previous Theorem 0.5 e f’(S)=2 f’(S)=1 Corollary. This is a 2-approximation algorithm for the minimum cost Steiner network problem. The residual problem is feasible.

25. First Try Observation : Half edges are good for degree bounds as well. Initialization: H =, f’ = f. While f’ ≠ 0 do: oFind a basic optimal solution, x, of the LP with function f’. oAdd an edge with x(e) ≥ 1/2 into H. oUpdate f’: for every set S, set o Update degree bounds : Output H. set B v :=B v -1 if e is incident on v. The residual problem may not be feasible! 0.5 e B v =2 B v =1

26. First Try Observation : Half edges are good for degree bounds as well. Initialization: H =, f’ = f. While f’ ≠ 0 do: oFind a basic optimal solution, x, of the LP with function f’. oAdd an edge with x(e) ≥ 1/2 into H. oUpdate f’: for every set S, set o Update degree bounds : Output H. set B v :=B v -0.5 if e is incident on v. The residual problem is feasible. 0.5 e B v =2 B v =1.5 Problem: A half edge may not exist!

27. A Proof Sketch An edge of 0, delete it. An edge of 1, pick it. Tight inequalities all come from the connectivity requirements. A basic solution is the unique solution of m linearly independent “tight” constraints, where m is the number of variables in the LP.

28. A Proof Sketch Uncrossing technique : uncross “intersecting” sets A B A[BA[B AÅBAÅB Each tight constraint corresponds to a set of vertices in the graph.

29. A Proof Sketch If there are no half-edges, derive a contradiction by a counting argument. no intersecting tight sets => number of linearly independent tight constraints is small no half-edges => number of edges is large In a basic solution, number of linearly independent tight constraints = number of edges Laminar family 1/3 Contradiction!

30. The Difference But integrality is important in the counting argument. Uncrossing would just work fine. fractional values B v =0.5 0.25 Back to our current problem. Why a half edge may not exist with fractional degree constraints?

31. New Idea Idea: Relax the problem by removing the degree constraint for v if v is of “low” degree. Intuition: Removing a constraint decreases the number of linearly independent tight constraint, and makes the counting argument work. Reason: Only violate the degree bound by an additive constant. B v =0.5 0.25 Lemma [LNSS]: If every vertex is of degree 5 when its degree constraint is present, then there is a half edge in a basic solution.

32. Iterative Relaxation Initialization: H =, f’ = f. While f’ ≠ 0 do: oFind a basic optimal solution, x, of the LP with function f’. o (Rounding) Add an edge with x(e) ≥ 1/2 into H. o (Relaxing) Remove the degree constraint of v if v has degree ≤ 4 oUpdate f’: for every set S, set o Update degree bounds: set Bv:=Bv-0.5 if e is incident at v. Output H. An additive constant +3 A multiplicative factor 2 Main Theorem [LNSS]: There is a (2,2B+3)-approximation algorithm for the minimum Steiner network problem with degree constraints.

33. Concluding Remarks [LS] There is a (1, B+1)-approximation algorithm for degree bounded minimum spanning tree problem. There is a (2, B+9)-approximation algorithm for the minimum Steiner forest problem with degree constraints. There is a (2, B + 6r max +3)-approximation algorithm for the degree bounded minimum Steiner network problem. Main Theorem [LNSS]: There is a (2,2B+3)-approximation algorithm for the minimum Steiner network problem with degree constraints. Question : Is there an additive approximation for this problem?

34. Concluding Remarks The iterative relaxation method provides a unifying algorithmic framework for many network design problems. And even for many polynomial time solvable problems. This method can be applied to other approximation algorithms. e.g. prize collecting Steiner trees, multi-criteria spanning trees, partial covering problem, degree-constrained graph orientations, etc. e.g. matchings, matroid intersection, submodular flows, Frank-Tardos, etc. A course is prepared by Mohit Singh on the topic.

35. A big thanks!