CS 3150 Presentation by Dan Li Advised by Kirk Pruhs

1. Primal-Dual Meets Local Search: Approximating MST’s with Non-uniform Degree BoundsAuthor: Jochen KönemannR. RaviFrom CMU CS 3150 Presentation by Dan Li Advised by Kirk Pruhs Department of Computer Science, University of Pittsburgh December 2, 2003

2. Motivation • Multicasting: • Nodes are connected by network. • Multicasting from one node to all other nodes • Cost associated to each connection • Cost effective solution • Minimum Spanning Tree’s Picture copied from the author’s talk

3. Motivation cont. • Problem: • Congestion: • Some nodes may be too busy to work effectively • Bandwidth limit • Solution • Bound the maximum number of connection that each node can support • Uniform bounds • Non-uniform bounds Picture copied from the author’s talk

4. Motivation cont. Picture copied from the author’s talk

5. Problem Formulation • Degree-bounded minimum-cost spanning tree problem with non-uniform degree bounds (nBMST). • Given an undirected graph G = (V, E), a cost function c : E → IR+ and positive integers all greater than 1, the goal is to find a spanning tree T of minimum total cost such that for all vertices the degree of v in T is at most Bv. • If all the Bv’s are the same, we have Degree Bounded Minimum-cost Spanning Tree Problem with Uniform Degree Bounds. • This problem is NP hard!

6. What is done in this paper • A new algorithm: • Improved approximation algorithms for the minimum cost degree bounded spanning tree problem in the presence of non-uniform degree bounds. • Direct algorithm, do not solve linear programs. • The algorithm integrates elements from the primal-dual method for approximation algorithms for network design problem with local search methods for minimum-degree network problem. • Goes through a series of spanning trees and improves the maximum deviation of any vertex degree from its respective degree bound continuously.

7. Core Theorem • Theorem 2: There is a primal-dual approximation algorithm that, given a graph G=(V, E), a nonnegative cost function c: E→IR+, integers Bv > 1 for all and a parameter ω >1, computes a tree T such that • It is apparent that • And the approximation ratio is constant. • More specifically, if we select b = 2 and ω = 2, we have

8. Primal-Dual formulation

9. High level idea • Intuition: • Reduce the degree those nodes whose degree is substantially higher than their bound Bv. • As we proceed through this sequence, while keeping the cost of the associated primal solution (tree) bounds with respect to the corresponding dual solution. • Define Normalized degree: ndegT (v) = max{0, degT(v) – βv·Bv} • Where βv > 0 are constants for all v in V. • How to choose βv? We will talk about it soon.

10. High Level Idea • Computer a sequence of MST’s (x1, {y1, λ1}), (x2, {y2, λ2}), …, (xt, {yt, λt}) • Until there is no such a node v with ndegT(v) ≥ 2 logb(n) • What is the difference between each computation? • On each re-compute step, raise the λ value of a carefully chosen set Sdof nodes with high normalized degree. Thus introducing more slacks. • Rerun the MST, taking advantage of the newly created slacks. • Also, keeping the cost close to the dual : Guarantee the approximation factor • Number of re-compute is polynomial : Guarantee it is a polynomial algorithm • If we look at the dual problem, we can intuitively consider usingCuv + λu + λvas the new cost function.

11. High Level Idea • What we are expecting? • By raising the value of λ’s, in the new MST‘s, some edges to/from the congested vertices can be replaced by edges between other nodes, thus decrease the normalized degree. • How to make this happen? • If some edges becomes more expensive, then it will be less preferred in MST. • If those edges to/from the congested node, then the congested node will be less preferred.

12. Visualization I

13. Visualization II

14. High Level Idea • How much do we increase the price • We expect that by increasing the price, there is only one edge difference between the old MST and the new MST. • We want to lose customer one by one • Increase too fast is bad, too few may not change the MST. • We do not want to lose all the connections to/from an edge, but only want to decrease the normalized degree to some controllable value. • Whose price to increase? • Only those edged connected to congested nodes

15. High Level Idea • How to end the process? • It may be difficult or impossible to decrease the normalized degree of each nodes to 0, which means we find a solution satisfying the bounds. • It may be feasible to decrease the normalized degree to some predefined level, then we find an algorithm that gives results that do not violate the bound too much. • The algorithm should end in polynomial number of steps. • Does such an algorithm exist?

16. The Algorithm

17. Analysis of the Algorithm • Initialize the primal-dual solution • Primal infeasible and dual feasible solution • Improve the primal feasibility and dual optimality • Some lines need to be clarified • Line 4 : Ends the algorithm • Line 5 : Used to select the set to increase the cost • Line 6 : How much to increase ? εi • Line 7 : Update the dual solution. • Line 8 : Update the cost function to re-compute the new primal solution • More questions: • How are the approximation factor are guaranteed? • How are the bounds satisfied (with linear factors)?

18. Clarifications • Line 4: On finishing ndegT (v) = max{0, degT(v) – βv·Bv} ≤ 2 logb(n) So, degT(v) ≤ βv·Bv + 2 logb (n) • Line 5: Selecting the set to increase the cost • Use contradiction, assume that no such di exists, and also consider that Bv ≤ n – 1 and

19. Clarifications • Line 6: Choosing εi • Such that the following run of MST yields a new tree that differs from the previous one by a single swap. • Cross-edge: e = uv is a cross-edge if • E is a non-tree edge, and • Where Ki is connected components of the forest • Choose: • And final εi to be the minimum among all the εie’s • The new MST must be different from the previous one, since we can swap one edge to form a new spanning tree with lower or equal cost than the one with previous selected edges • Local improvement

20. Performance Analysis • The cost is close to the dual cost by a constant factor • On each step, we need to maintain the cost to be close to the dual cost • The optimal solution dual solution is the optimal primal solution, so dual solution is less than the optimal • Thus the relation between the primal cost and the optimal is maintained • On each iteration, yπ’s may increase and λπ may also increase, and also the spanning tree cost. • The first term on the right-hand side should grow sufficiently to compensate for the decrease in the second term and also increased spanning tree cost.

21. Performance Analysis • In order to prove the previous equation, an invariant is proved. • Induction • Base case i = 0 • Induction • Selecting ` , (Inv) is proved.

22. Performance Analysis Plus this • Following equations can be reached (see the paper for details) • Concludes ( by choosing α ≥ ω )

23. Analysis - Running time • This algorithm terminates in polynomial number of steps • Claim: Algorithm 1 terminates after O(n4) iterations? • Proof: Define the potential of spanning tree Eias On each step, one edge is swapped in, which is incident to two nodes of normalized degree at most di - 2. The reduction of the potential is at least

24. Analysis - Running time • Consider that • The equation on the last page is bounded by • Also consider that the initial potential Φi at the beginning of ith step is at most , after the ith step, or at the beginning of the (i+1)th step, the potential Φi+1 is at most • With O(n3) iterations, the potential function is reduced by a constant factor. • The algorithm runs for O(n4) iterations total??? • Considering that each iteration can be implemented in time O(n2log(n)), the whole algorithm runs in time O(n6log(n))

25. Is the analysis correct? • The above analysis appears in the paper, is it correct? Look at this • If b = 3, the left side is • If b = 9, the left side is • If b = 2, the left side is • So the correctness of the above equation is dependent on the value of b. • Only when b >= 3, the running time is O(n6log(n)) • In the recent talk given by the author, he used value of b as 2, so the analysis is wrong.

26. More Problems? • Is there anything missed? • Did the author prove the part 1 of theorem 2? • No. • It seems apparent, since on finishing the while loop, the maximum normalized degree is 2 log(n), then • But βv is selected as • Which can not continue to prove

27. More Problems? Solve? • The conclusion can still be correct if we selected special value of ω = 2, and • What value can be used for b? • Any value larger than 1 can be used • But only value larger than or equal to 3 can give running time of O(n6log(n)). • Smaller value of b will give worse running time.

28. Conclusion • The performance of the algorithm is conditional based on the value of constants selected. • What we learn from this paper? • Modify the cost function to avoid congestion • This is a very naturally and decent solution.