Primal-Dual Meets Local Search: Approximating MST’s with Non-uniform Degree Bounds Author: Jochen Könemann R. Ravi From CMU CS 3150 Presentation by Dan Li Advised by Kirk Pruhs Department of Computer Science, University of Pittsburgh December 2, 2003

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

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

Motivation cont. Picture copied from the author’s talk

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 B v.  If all the B v ’s are the same, we have Degree Bounded Minimum- cost Spanning Tree Problem with Uniform Degree Bounds.  This problem is NP hard!

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.

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 B v > 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

Primal-Dual formulation

High level idea Intuition:  Reduce the degree those nodes whose degree is substantially higher than their bound B v.  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: ndeg T (v) = max{0, deg T (v) – β v ·B v } Where β v > 0 are constants for all v in V. How to choose β v ? We will talk about it soon.

High Level Idea Computer a sequence of MST’s (x 1, {y 1, λ 1 }), (x 2, {y 2, λ 2 }), …, (x t, {y t, λ t })  Until there is no such a node v with ndeg T (v) ≥ 2 log b (n)  What is the difference between each computation? On each re-compute step, raise the λ value of a carefully chosen set S d of 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 using C uv + λ u + λ v as the new cost function.

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.

Visualization I

Visualization II

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

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?

The Algorithm

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 ? ε i  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)?

Clarifications Line 4: On finishing ndeg T (v) = max{0, deg T (v) – β v ·B v } ≤ 2 log b (n) So, deg T (v) ≤ β v ·B v + 2 log b (n) Line 5: Selecting the set to increase the cost  Use contradiction, assume that no such d i exists, and also consider that B v ≤ n – 1 and

Clarifications Line 6: Choosing ε i  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 K i is connected components of the forest Choose:  And final ε i to be the minimum among all the ε i e’ 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

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, y π ’s may increase and λ π 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.

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

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

Analysis - Running time This algorithm terminates in polynomial number of steps  Claim: Algorithm 1 terminates after O(n 4 ) iterations?  Proof: Define the potential of spanning tree E i as On each step, one edge is swapped in, which is incident to two nodes of normalized degree at most d i - 2. The reduction of the potential is at least

Analysis - Running time Consider that  The equation on the last page is bounded by  Also consider that the initial potential Φ i at the beginning of i th step is at most, after the i th step, or at the beginning of the (i+1) th step, the potential Φ i+1 is at most  With O(n 3 ) iterations, the potential function is reduced by a constant factor.  The algorithm runs for O(n 4 ) iterations total???  Considering that each iteration can be implemented in time O(n 2 log(n)), the whole algorithm runs in time O(n 6 log(n))

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(n 6 log(n)) In the recent talk given by the author, he used value of b as 2, so the analysis is wrong.

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 β v is selected as Which can not continue to prove

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(n 6 log(n)).  Smaller value of b will give worse running time.

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.