Improved Approximation for Minimum Power covering problems G. Kortsarz

1. Improved Approximation for Minimum Power covering problemsG. Kortsarz Joined work with Calinescu and Nutov

2. A description of IRR • Write a Set Cover like LP that seeks to solve some Covering Problem using a union of a collection of subsets. • Full Steiner tree: all leaves are terminals. Zelikovsky: Use trees with k terminals. • Decompose a tree to (non disjoint) full Steiner k-trees.

3. Decomposition into non disjoint full Steiner trees • Using non disjoint full Steiner 3-trees leads to a penalty of 1.6666… in the ratio. • For k=3 there is a PTAS to approximate the best 3-Trees cover. • This easily gives 1.66667 ratio.

4. Reduction in the optimum value • Byrka, Grandoni, Rothvo and Sanita in their seminal paper used the Bi-directed Cut Relaxation. And the Directed components cut relaxation. • The sets in the LP are the directed components. • That is the Set Cover like way need to cover the terminals.

5. Remark on their LP • Number of variables and constrains exponential but only small components are taken. • Its good to know that in general exponential constrains and variable may not be a problem as you can project the variables into a convex function.

6. How the ratio for Steiner tree improved • Ratio2(folklore?) • 1.83 [Zelikovsky ’93] • 1.667[Pr ̈omel & Steger 97] • 1.644[Karpinski Zelikovsky ’97] • 1.598[Hougardy & Pr ̈omel ’99] • 1.55[Robins & Zelikovsky 00] • Byrka et al ln 4<1.39 using LP.

7. Reduction in the optimum • In Set Cover, a combinatorial analysis of Greedy, might as well assume ROPT=, for the solutionRso far. To get a constant ratio we must show that the optimum goes down. • In this paper they show that the (expected) cost of the residual problemgoes down.

8. An Intro into a different objective: not the sum of edge costs • In a Wired Network each edge chosen has to be installed and we need to pay for it. • What if we deal withWireless Network? Of course we discuss the static case only, which is basicallySensor Network

9. Motivation-Wireless Networks • Nodes in the network correspond to transmitters • More power  larger transmission range • transmitting to distance r requires rpower, 2  4 • Transmission range = disk centered at the node • Battery operated  power conservation critical • Type of problems: • Find min-power range assignment so that the resulting • communication network satisfies prescribed properties.

10. The power of a vertex and a graph pE'(v)=Max{eE' touching v}{c(e)} • Many classical problems have been studied with respect to the (more difficult) min-power model • P(G)=v p(v). The sum over each vertex of its largest cost edge.

11. e e d d f f c c g g b b a a Range assignment Communication network

12. EXAMPLE UNIT COSTS c(G) = n p(G) = 1 c(G) = n p(G) = n + 1

13. Example • Example of power versus cost. A B C 6 5 3 5 8 6 D F 7 6 5 J H 6 P(A)=7, P(B)=8, P(C)=6, P(D)=6, p(F)=8, p(H)=6, p(J)=7 P(G)=7+8+6+6+8+6+7=48. C(G)=57

14. Comparing Power And Cost: Spanning Tree Case Min-cost version is a polynomial problem • Min-power spanning tree: is NP-hard[Clementi, Penna, Silvestri, 2000] • Best known approximation ratio: 5/3 [E. Althaus, G. Calinescu, S.Prasad, N. Tchervensky, A. Zelikovsky, 2004]

15. The Min-Power Vertex k- Connectivity Problem • We are given a graph G(V,E)edge costs and an integerk • Design a min-power subgraphG(V, E) so that every u,v Vadmits at least k vertex-disjoint paths from utov • May seem unrelated to Min cost vertex k-connectivity. But equivalent w.r to approximation (difficult proof).

16. The problem we study • Only undirected graphs • The Minimum Power Terminals cover: • Input: A graph G(V,E) c:EN^+, and a set TV of Terminals • Output: A minimum power graph so that each terminal has degree at least 1.

17. The Set Cover we choose • We reduce to the problem of covering the terminals with stars with at most k leaves with ksome constant. • In [KN] it is shown that the penalty is only1+1/k. And is negligible for large constant k. • Can avoid by using a separating oracle for general stars

18. Our paper technically is not exactly IRR • TheSet Cover LPis covering terminals with stars with a constant number k of leaves. • Such a star S selected randomly using the LP. • Since k is constant we can take the OPTIMUM COVER for the leaves of S. We DO NOT choose S.

19. What is known about covering k terminals for a constant k • As far as I know the best running time to solve this problem of Coveringk Terminals with Minimum Power is 2k∙poly(n). • Thus for constant k negligiblepenalty and polynomial time.

20. We do not show a reduction in the optimum • Motivated by a paper of Grandoni we define an interesting Potential function (R) with R the optimal solution for the remaining terminals . We show that our power is at most p(R)+ (R). • The proof shows that (R) goes down at every iteration.

21. Power problems are harder • For sum of costs it’s a matching-like problem and so can be solved in polynomial time. • For Minimum power NP-Hard, even if the terminals T are an independent set. • Minimum Power NP-hard for uniform costs.

22. Our Results • For the general Minimum Power Terminals Cover we give a 1.41<1.5 ratio using IRR. • If T is an independent set the head is not a terminal, thus we get a 1.22 ratio using IRR. • For uniform cost, 1.2 ratio.

23. Finally, The Backup Terminals Problem • We study the following problem: Choose a Minimum Power Subgraph H So That For Every Terminal t There is in H a Path to Another Terminal t’. • There is a trivial 2ratio. • We give a 1.5ratio.

24. Selecting edges so that Every terminal has degree at least 1 • The solution is a collection of stars. • The only difference is the root. If the cost of the edges in S are ck≤ ck-1 ≤ ck-2≤…… ≤ c1 p(S)= 2∙c1+ c2+ c3+ ck It is interesting that counting c1 twice makes such a difference.

25. Example: the unfilled nodes are terminals. A minimum power cover 4 1 2 1 2 5 1 7 3 2 6 3 8 3

26. Example: the unfilled nodes are terminals. A collection of stars 1 2 1 1 3 3

27. The most related to us result • Grandoni gives a 1.91 ratio algorithm for The Minimum Power Steiner Tree. • We use the same potential function. But in different ways. • Grandoni covers cuts, we cover terminals. • Combined with an algorithm by [KN]. IRRalone gives nothing

28. ASet-Coverlike LP with k-Stars • Minimize S pS ∙xS • Subject to tS xS ≥1 xS ≥0 • The LP has only stars with k leaves (not mentioned again). • We prove xS ≤|T| ≤n

29. A sample space • The values {xS/n} are a sample space (polynomial size). 1 While there is non cover term’ 1.1 Write the LP. 1.2 Chose some S at random. 1.3 Find opt for S and add 1.4 Delete all terminals of S 2. Return the solution.

30. The probability that a terminal is covered • By the LP, a vertex is covered with probability 1/n. • Thus in O(n∙log n) iterations we are done • Our solution for covering the chosen S optimal hence better than fractional solution. • But less than the cost the optimum used to cover S.

31. The following proof works for any potential function • Let Ri-1 be the optimal stars of size k for terminals uncovered, before iteration i-1. • Ri results from Ri-1 by removing covered terminals in iteration i-1 • Elaborate later. Denote the potential function by (Ri)

32. The axiom needed • (Ri-1) - (Ri) ≥ p(Ri-1)/n • Let J be the solution that we output by our algorithm. This is the union of the optimal integral solutions over all stars chosen at all iterations. • We show that p(J)≤ (R0) • This can be proved for any 

33. Proof • Denote opti =SRi-1p(S) • The expected fractional cost when it ends is  opti /n • Thus  opti /n≤ p(Ri-1)/n • But in IRRJ, takes the optimum for each chosen star. • p(J) ≤  opti /n

34. Proof • p(J) ≤  opti /n ≤  p(Ri-1)/n≤  ((Ri-1) - (Ri))= (R0) • The algorithm itself takes the minimum between the above solution and some solution by [KN]. More precisely a convex combination.

35. The potential function • (R)=i≥1ci/i • The bound on opt is (R)=(R)+p(R). • The value of the solution of [KN] • p(R)+c3+c4+…….+cq • In[KR] it is shown that choosing 2-stars gives 3/2 ratio.

36. Different behavior • The ratio of (R)=i≥1ci/i over ci c1+i≥1ci goes down as q goes up. • For small q, [KN]is good, as c1and c2 start to be dominating. • We take a convex combination, with the [KN] getting 2/3 and the IRR gets a 1/3. The optimization is unusually hard.

37. Inequality I • What happens if we add a collection of edges H. Let the reduction in the potential be denoted by (H). • How doess this compare to  vH (v)? • It turns out that (H)≥  vH (v).

38. A technically complex proof • True only for the (H) we defined. • But now we explain how we finish the proof of the main lemma. • And only in the case the head is not a terminal. If it is a terminal very complex.

39. Proof in two parts • Denote by Pr(H) the probability exactly H is hit. • Exp(())=Pr(H)∙ (H). • Indeed, for ever H that can be hit, it will bring (H) change as by notation and the probability for that is that H is hit, is denoted Pr(H).

40. Two inequalities • We will show two things: 1) Exp(())=Pr(H)∙(H)≥  vR (v)/n 2)  vR (v)≥p(R). 3) Together gives Exp(()) ≥p(R)/n

41. First inequality • Exp(())=  Pr(H)∙(H)≥  Pr(H)  vR (v) • This holds by Inequality 1. • ≥ vR (v)/n∙ Pr(v is hit). • The reason is that we add the probability of every H that contains v this is exactly the probability that v is hit.

42. Continued • vR (v) Pr(v is hit)≥ vR (v)/n • Recall that every vertex is hit with probability 1/n. • Note this this ends the proof of the first inequality. • The second inequality hard to prove. See the paper.

43. And this is only the easy case • A much more complex case is that the head of the star is a terminal. • One thing is clear: this paper is very very hard technically. A simpler proof will be wellcome but seems unlikely

44. Can we turn this into a collection of axioms that imply a good ratio? • Unfortunately many of the properties hold just because the specific potential function  chosen. • A nice (probably hard) question is can we give some axioms that imply some ratio? • Can we get something general?

45. Is there a lesson from this paper? • It does not seem so. Even though the same potential function appears in two papers. • I like general theorems that work for many problems like the GW primal dual and the Jain Result. • Otherwise its just one approximation after the other.

46. Open problems • There are very few results that use IRR. • The potential function seems to be an idea that can help. • Can we solve an LP once? • A non constant lower bound for Minimum Cost vertex k-Connected Subgraph?