PrimalDual Meets Local Search: Approximating MSTs with Nonuniform Degree Bounds Author: Jochen Knema

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Primal-Dual Meets Local Search Approximating
MSTs 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

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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 Trees

Picture copied from the authors talk
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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 authors talk
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Motivation cont.
Picture copied from the authors talk
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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 Bvs are the same, we have Degree
  Bounded Minimum-cost Spanning Tree Problem with
  Uniform Degree Bounds.
• This problem is NP hard!

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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.

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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
  gt 1 for all and a parameter ? gt 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 ? 2,
  we have

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Primal-Dual formulation
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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) max0, degT(v) ßvBv
• Where ßv gt 0 are constants for all v in V.
• How to choose ßv? We will talk about it soon.

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High Level Idea

• Computer a sequence of MSTs (x1, y1, ?1),
  (x2, y2, ?2), , (xt, yt, ?t)
• 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 ? value of a
  carefully chosen set Sd 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 usingCuv ?u ?v as the
  new cost function.

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High Level Idea

• What we are expecting?
• By raising the value of ?s, in the new MSTs,
  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.

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Visualization I
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Visualization II
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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

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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?

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The Algorithm
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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 ? ei
• 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)?

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Clarifications

• Line 4 On finishing
• ndegT (v) max0, degT(v) ßvBv 2 logb
  (n)
• So,
• degT(v) ßvBv 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

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Clarifications

• Line 6 Choosing ei
• 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 ei to be the minimum among all the
  eies
• 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

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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, yps may increase and ?p 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.

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Performance Analysis

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

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Performance Analysis

• Following equations can be reached (see the paper
  for details)

Plus this

• Concludes ( by choosing a ? )

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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 Ei as
• 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

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Analysis - Running time

• Consider that
• The equation on the last page is bounded by
• Also consider that the initial potential Fi at
  the beginning of ith step is at most ,
  after the ith step, or at the beginning of the
  (i1)th step, the potential Fi1 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))

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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 gt 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.

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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 ßv is selected as
• Which can not continue to prove

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More Problems? Solve?

• The conclusion can still be correct if we
  selected special value of ? 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.

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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.