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Approximation Algorithms

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Approximation Algorithms

Department of Mathematics and Computer Science

Drexel University

- We are given a graph G=(V,E) with edge weights and a set of vertices selected out of V say pairs si,ti such that si and ti need to communicate. We'd like to find a minimum cost sub-graph such that all the <si,ti> pairs are connected.
- Given: G=(V,E) w.l.o.g. fully connected,
- weights we>= 0, for every edge e in E
- <s1,t1>, <s2,t2>,â€¦, <sk,tk> paris of communicating edges

- Find: A minÂcost subgraph H of G such that for all values of i in {1,â€¦,k} the pairs <si,ti> â€™s are connected in H.

- Define L(S) as the indicator of subset S of V if S forms a cut that separates one of the <si ,ti> pairs. More formally:
- Also, for a graph G=(V,E) and a subset of vertices S defined(S)={e=(u,v) in E : u in S and v not in S}. That is, d(S) is the set of edges that cross the cut boundary.

- Let X=(x1,â€¦, xm) be a feasible solution for previous ILP, and assume to the contrary that there exist a communicating pair (u,v) that is not connected using the edges in set {e| xe=1 (the entry e in X)}.
- There are many cuts that separate u and v. Let S be once such cut that contains u. We will also add all the vertices in the same connected component as u to S.
- Since u and v are disconnected with respect to edges selected by X. Sum of the xe with respect to cut S is zero, but L(S)=1, and the cut edges violate the constraint of cut S, that contradicts the feasibility of X.

- We will relax the integrality condition to xe>=0. As a result:

- A dual variable is raised if ys>0.
- A set S of vertices is unsatisfied if L(S)=1 but no cut edge between S and V-S has been selected by algorithm yet.
- The smallest unsatisfied set is referred to as active set.
- Edge e is said to be tight if:

- The algorithm starts with no edges picked and no cuts raised.
- During any iteration of the algorithm:
- the current primal solution indicates which sets should be raised,
- and the current dual solution indicates which edge needs to be picked.

- Thus, the algorithm iteratively improves both the primal and dual solution integrally until the primal solution becomes feasible.

- Let us describe what happens in an iteration.
- Let F denote the sub-graph the algorithm is constructing.
- Using the primal solution, the algorithm finds all the active sets in F and raises the dual variables of these sets simultaneously until some edge goes tight.
- Note that it is the dual solution at the beginning of the iteration that determines how high the dual variables of the active sets can be raised. Any one of the tight edges is picked and added to F.

- The edge augmentation step terminates once F becomes a feasible primal solution.
- Since F may contain redundant edges, that is edges whose deletion still makes F feasible, a final pruning step is necessary to achieve the desired approximation factor.

- before proving the correctness of the algorithm, we will show that it is not difficult to find all the active sets in an iteration. In fact, it is sufficient to consider all the connected components of the current F .
- Set S is active at iteration i if and only if it is a connected component of F (at iteration i) and L(S) = 1:
- if Set S is active then L(S) = 1 and is minimal, and hence connected in the current sub-graph F .
- Next, if S is a connected component of F and L(S)=1, then S is a minimal unsatisfied set, otherwise, there is a larger connected component of F that contains S.

- At the end of the algorithm, Fâ€™ and y are feasible primal and dual solutions:
- At the end of the edge augmentation step, F satisfies all the connectivity requirements. Furthermore, it is not difficult to see that F is a forest, since only the dual variables of connected components are raised. That is, none of the edges in a connected component will become tight in later iterations. Since F is a forest, if u and v are a communicating pair, then there is a unique path from u to v. None of the edges along this path can be pruned and therefore, Fâ€™ remains primal feasible after the pruning step.

- Moreover:
- Suppose edge e becomes tight at an iteration. If e is picked, then all the sets S such that e belongs to d(S) with L(S) = 1 becomes satisfied and none will be active in later iterations. Consequently, all the dual variables of these sets will never be raised in the future and the constraint corresponding to edge e remains valid. But suppose edge e is not picked at the end of the iteration nor at succeeding iterations before the algorithm raises the dual variables of some active sets. This implies that the sets S such that e belongs to d(S) with L(S) = 1 are no longer active. We follow the same argument above that the constraint corresponding to e in the dual remains valid at the end of the algorithm.

- Let degFâ€™(S) denote theedges inFâ€™that cross the cut (S,V-S) then:
- .

- In fact, we can prove a stronger result: that in each iteration, the increase in the l.h.s. of this inequality is bounded by the increase in the r.h.s. of the inequality. Consider an iteration. Let D be the increase in the dual variables of the active sets in this iteration. Thus, we wish to show that:

- That is, we need to show that the average degree of the active sets with respect to Fâ€™ is at most 2. Notice that these degrees are due to the edges that will be picked during or after the current iteration.
- The heart of argument relies on the fact that the average degree of a tree or a forest is at most 2.
- Consider the graph Fâ€™. Contract all the nodes in each active set S to a single node and let us call these nodes active.
- Delete all the singletons. Call this new graph H. Then the degree of the active node corresponding to set S in H is degFâ€™(S).
- Since Fâ€™ is a forest, H must also be a forest. To show that the average degree of the active nodes in H is at most 2, it is sufficient to show that all inactive nodes in H must have degree at least 2, since the average degree of all the nodes in a forest is at most 2.

- At iteration i, let C be a connected component w.r.t. F. If L(C) = 0 then degFâ€™(C) is not 1.
- Suppose there is a unique edge e of Fâ€™ that crosses the cut (C,V-C). Since Fâ€™ does not contain redundant edges, there must be a communicating pair u and v such that u belongs to C and v in V-C. This implies that L(C) = 1, leading to a contradiction.

- Thus, if l in H is not active, l must be part of some connected component w.r.t. F at iteration i, say C, and L(C)=0. Furthermore, if l is a leaf in H then degFâ€™(C)=1 contradicting the claim above. Thus, all leaves in H are active and all inactive nodes in H have degree at least 2.

- Initialization:

- Communicating pairs:
<u,v>, <s,t>.